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Kci*fTOi
Di'ized by Google
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THE LONDON SCIENCE CLASS-BOOKS
ELEMENTARY SERIES
EDITED BY
PROF. G. C. FOSTER, F.R.S. and SIR PHILIP MAGNUS
HYDROSTATICS AND PNEUMATICS
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HYDROSTATICS AND PNEUMATICS
BY
SIR PHILIP MAGNUS
OF THE CITY AND GUILDS OF LONDON INSTITUTE ;
AUTHOR OF 'lessons IN ELEMENTARY MECHANICS/ &C
JOINT EDITOR OF THIS SERIES
EIGHTH EDITION
LONDON
LONGMANS, GREEN, AND CO.
AND NEW YORK: 1 5 EAST 1 6* STREET
189I
All rights reserved
Digitized by VjOOQIC
/{c»4lo I
HARVARD
UNIVERSITY
LIBRARY
' ^92
^
Richard Clay & Sons, Limited,
London & Bungay.
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EDITORS' PREFACE.
Notwithstanding the large number of scientific
works which have been published within the last few
years, it is very generally acknowledged by those who
are practically engaged in Education, whether as
Teachers or as Examiners, that there is still a want of
Books adapted for school purposes upon several
important branches of Science. The present Series
will aim at supplying this deficiency. The works
comprised in the Series will all be composed with
* special reference to their use in school-teaching ; but,
at the same time, particular attention will be given
to making the information contained in them trust-
worthy and accurate, and to presenting it in such a
way that it may serve as a basis for more advanced
study.
In conformity with the special object of the Series,
the attempt will be made in all cases to bring out the
educational value which properly belongs to the study
of any branch of Science, by not merely treating of its
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VI Editors' Preface,
acquired results, but by explaining as fully as possible
the nature of the methods of inquiry and reasoning by
which these results have been obtained. Conse-
quently, although the treatment of each subject will
be strictly elementary, the fundaro'^ntal facts will be
stated and discussed with the fulness needed to place
their scientific significance in a clear light, and to
show the relation in which they stand to the general
conclusions of Science.
In order to ensure the efficient carrying-out of the
general scheme indicated above, the Editors have
endeavoured to obtain the co-operation, as Authors
of the several treatises, of men who combine special
knowledge of the subjects on which they write with
practical experience in Teaching.
The volumes of the Series will be published, if
possible, at a uniform price of \s, 6d, It is intended •
that eventually each of the chief branches of Science
shall be represented by one or more volumes.
G. C. F
P. M.
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PREFACE.
This Class-Book is intended for the use of those
pupils in the upper forms of schools who have
already acquired some elementary knowledge of the
principles of Mechanics, and are about to commence
the study of Hydrostatics and Pneumatics.
In the treatment of the subject of this volume I
have endeavoured, as far as possible, to combine the
Experimental with the Deductive method. When-
ever a law is stated, some explanation is afforded of
the several experiments by which that law has been
established ; and whenever a result is deduced, by
the aid of mathematical reasoning, from more ele-
mentary principles, the pupil is shown how this
result may be experimentally verified
In the hope that this Uttle work may serve as an
introduction to more advanced treatises on Hydro-
statics, I have devoted a few pages to the consider-
ation of the * Flow of Liquids through Pipes and
Small Orifices'; and, whilst avoiding the mathe-
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yiii Preface.
matical difficulties which the fuller treatment of this
branch of the subject involves, I have endeavoured
to bring into prominence some of the leading prin-
ciples connected with it, which recent investigations
have aimed at establishing.
To facilitate the use of this text-book in class-
instruction, the subject-matter is divided into a
number, of short sections, in which all the more im-
portant propositions are illustrated by numerical
examples. To nearly every section is appended a
set of exercises, progressively arranged, to be solved
by the pupil.
My obligations to the published works of different
writers are acknowledged in the body of the book.
P. M
London^ Savile Club :
September 1878.
IMs volume can be obtained with or mthout the answers to the
Exerciser
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CONTENTS.
CHAPTER I.
PRELIMINARY NOTIOKS.
'.ECTION PAGR
I. Nature of Fluid Bodies I
Essential Diflferences between Solid and Fluid
Bodies. — Change of Gaseous into Liquid State. —
Experiments of Dr. Andrews. — Viscosity. — The
Three States of Matter.
II. Units of Measurement . . . . . . 8
Units of Length, Area, and Volume. — C.G.S.
System of Units. — Units of Mass and Weight,
— Derived Units. — Geometrical Relations.
Ill Density, — Specific Gravity . . . ^ , 12
Definitions and Measures of Density and Spe-
cific Gravity. — Ordinary Definition of Specific
Gravity. — Tables of Specific Gravity. —Density
and Specific Gravity of Compound Substances. —
Exercises.
CHAPTER IL
FLUID PRESSURE ON SURFACES IMMERSED.
IV. Explanation of Terms. — FascaTs Principle . 20
Intensity of Fluid Pressure. — Variation of Pres-
sure with Depth. — Direction of Pressure. — Equal
Transmissibility of Fluid Pressure. — Mechanical
Appliances — Communicating Vessels. — Exer-
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X Contents^
SECTION PAGf
V. Whole Pressure on Surface Immersed . . .30
Definition and Measure of Whole Pressure. —
Pressure on Base and Sides of a Vessel containing
Liquid. — Pascal's Vases, — Exercises.
VI. Centre of Pressure -37
Definition. — Centre of Pressure of Rectangular
Area — Of Triangle.— General Expression. —
Examples. — Exercises.
CHAPTER III.
FLUID PRESSURE ON BODIES IMMERSED.
VII. Resultant Vertical Presmte. — Principle of Arcki-
medes 44
Determination of Resultant Vertical Pressure. —
Principle of Archimedes. — Experiments. — Real
and Apparent Weight of Bodies. — Relative Gra-
vitation of a Body in a Fluid. — Exercises.
VIII. Floating Bodies, — Metacentte . . . . £i
Principle of Flotation. — Stable and Unstable
Equilibrium. — Definition of Metacentre. —
Examples. — Exercises.
CHAPTER IV.
SPECIFIC GRAVITY, AND MODES OF DETERMINING IT.
IX. Application of Principle of Archimedes to the deter-
mination of the Specific Gravity of Bodies . 61
Specific Gravity of a Heavy Solid insoluble in
Water.— Of a Solid that Floats in Water.— Of
a Liquid.— Of a Solid soluble in Water.— Of
Gases. — Exercises.
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Contents. xl
SECTION PAGE
X. Other Methods of Determining Specific Gravity.'^
Hydrometers ...... 66
The Specific Gravity Bottle.— Fahrenheit's Hy-
drometer.— Nicholson's Hydrometer. — Exer-
cises.
CHAPTER V.
THE MOTION OF LIQUIDS.
XI Liquids momng by their own Weight 7 1
Definition of Pressure Height or Head of Liquid.
— Torricelli*s Theorem. — Relation of Velocity
of Flow to Sectional Area of Vessel. — Vena
Contracta. — Liquid Flowing through a Pipe of
Variable Area. — Relation of Velocity of Flow
to Pressure. — Flow of Liquid through Small
Orifice. — Effect of Friction on Pressure at Diffe-
rent parts of a Pipe.
Xil. Capillarity 86
Drop Formation. — Experiments. —Surface Ten-
sion. — Capillary Phenomena, Experiments.--
Principles of Capillarity. — Law of Diameters.
XIII. Diffusion of Liquids 94
Graham's Experiments. —Crystalloids and Col-
loids. — Dialysis. — Osmose. — Structure of Li-
quids.
CHAPTER VI.
THE PRINCIPLES OF PNEUMATICS.
XIV. General Properties of Gases, — Atmospheric Pressure 102
Definition of Pneumatics. — Expansibility and
Compressibility of Gases. — Experiments showing
that the Air has Weight. — Measure of Atmo-
spheric Pressure. — Torricelli's Experiment. —
Barometers. — Barometric Corrections. — Abso-
lute Pressure per Umt Area. — The Siphon. —
Exercises.
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xii Contents.
SECTION ^^®^'
XV. BoylisLaw . . . • • * * "4
Experiments for Pressures greater than Atmo-
spheric Pressure.— Results of Expenments.—
For Pressures less than Atmospheric Pressure.
— Dalton's Law. — Graphic Representation of
Boyle's Law.— Limits of Boyle's Law; Expen-
ments of Regnault and Despretz.— Relative
Densities of the Air at different Heights.—
Use of Barometer in Determining Heights.—
Exercises.
XVI. Diffusion of Gases '^9
Experiments. — Rate of Diffusion. — Graham's
I^w.— Kinetic Theory of Gases.
CHAPTER Vn.
PNEUMATIC INSTRUMENTS.
XVn. Diving Bell— Pressure Gauges
I'^e
141
Description of and Problems on Divmg Bell.
—Manometers.— Compressed Air Manometer.
—The Siphon Gauge.— Exercises.
XVni. Air-Pumps
Parts of a Pump.— Single-Barrelled Air-Pump.
—Density of Air after a certain number ot
Strokes.-Difficulty of Working. -Double-
Barrelled Air- Pump. -Tate's Air-Pump.-
Magdeburg Hemispheres. - bprengels Air
Pump.-Condensing Syringe. —Exercises.
XIX. Pumps for Liquids 5
Common or Suction Pump. — Tension of
Piston-Rod.-The Lifting Pump^The For-
cing Pump. -With Air Vessel.— Fire-Engme.
Bramah's Press.— Exercises.
Miscellaneous Problems
162
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HYDROSTATICS.
CHAPTER I.
PRELIMINARY NOTIONS.
I. Nature of Fluid Bodies,
§ I. Hatter. — What we call matter may exist in
the solid form as iron, wood, and ice, or in the fluid
form as water, oil, air, and steam.
§ 2. Essentjial differences between Solid and
Fluid Bodies. — If we take any portion of a solid body,
such as a piece of metal, a sheet of glass, or a block of
wood, one of the first things we observe is that it
possesses a definite shape, which cannot be changed
except by the application of a certain amount of force ;
we may also observe that it occupies, wherever it may
be placed, the same amount of space. If, however,
we take a given portion of a fluid substance, such as
water or air, we find that it possesses no definite shape,
and that it moulds itself to the form of the vessel in
which it is contained. Thus if we pour a certain
quantity of water from one vessel into another, we
observe that whilst the volume of the liquid remains
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2 Hydrostatics,
the same, its shape changes with that of the portion
of the vessel which it occupies.
If, again, we endeavour, as in the act of cutting,
to separate one part of a solid from another, a certain
amount of pressure, depending on the nature of the
material, must be exerted ; but if we pass a smooth
plane surface, such as the blade of a knife or a sheet
of glass, in the direction of its plane, through a mass
of fluid, very little resistance is experienced.
These experiments show that the particles of a
fluid are mobile^ i.e., they move freely among one
another, and cohere so feebly that they can be
separated from oae another by the application of a
very slight force. We may, therefore, define a perfect
fluid as a substance the particles of which move freely
among one another.
If we take a straight cylindrical lube, open at one
end and closed at the other, and fit into it a smooth
rod of some solid substance, such as iron or wood,
and press the free end of the rod in the direction of
its length, the pressure applied is transmitted to the
closed end of the tube without producing any effect
on its curved surface ; but if the tube contain a fluid
instead of a solid, and pressure be applied to it by
means of a piston fitting into the tube, the pressure
will be felt not only at the base of the vessel, but like-
wise at all points in its curved surface. This difference,
which is characteristic of these two states of matter,
is also due to the fact that the particles of a solid,
being more or less rigidly connected, are capable of
holding together under the influence of a certain
amount of force, whilst those of a fluid, being to a
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Liquids and Gases. 3
great extent independent of one another, tend to move
away in all directions when acted upon by a force in
any one direction. For this reason fluid bodies under
the action of gravity cannot rest on a hard horizontal
surface, otherwise unsupported, as solids do, but re-
quire in addition lateral support. Hence the follow-
ing definition : FMid bodies are those which cannot
sustain a longitudinal pressure, however small, without
being supported by lateral pressure also.
§ 3. Two kinds of Fluids. — There are two great
classes of fluids, called liquids and gases, which are thus
distinguished : —
If a given quantity of water, which is liquid, be
poured into a vessel, it will occupy a certain portion of
the vessel and no more ; but if a small portion of
hydrogen or carbonic acid, which are gases, be intro-
duced into a vessel, however large, the gas will expand
so that some of it will be found in every part of the
vessel.
If we take a cylindrical vessel fitted with ^<^ »•
a piston and fill it with water, we shall find HT
that no amount of pressure we can apply j— 1~
will sensibly diminish the volume of the
water; but if, the vessel being filled with
air, we press down the piston, the volume
occupied by the gas will be found to
diminish as the pressure is increased.
Liquids, when submitted to very considerable pres-
sure, have been found to undergo some diminution of
volume, but so little, that for most practical purposes
they may be considered as incompressible fluids.
Hence a. perfect liquid may be defined as a mass
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4 Hydrostatics.
which is absolutely incompressible and absolutely
devoid of resistance to change of shape. Matter,
however, satisfying this condition does not exist in
nature.
The chief difference between the liquid and the
gaseous state of matter consists in this, that whilst a
given portion of a liquid has a definite volume, but no
definite shape, a given portion of a gas has neither defi-
nite volume nor shape, its volume and shape being
always the same as that of the vessel containing it
§ 4. Change from Oaseons into Liquid State. —
Some substances exist at ordinary temperatures both
in the liquid and gaseous state. Thus we have steam
and water, and ether both as a liquid and as a gas or
vapour. Other gases cannot be reduced to the liquid
state except under the influence of extreme cold and
great pressure. The temperature at which the change
takes place, and the amount of pressure required,
vary considerably. Until very recently, all efforts had
failed to reduce certain gases to the liquid condi-
tion ; and consequently these gases were called
permanent gases, as distinguished fi-om vapours or
liquefiable gases. But the experiments of MM. Pictet
and Cailletet, performed in December 1877, have
demonstrated as a fact what was previously only an
inference firom analogy, that every gas is the vapour
of some liquid, and can be reduced to a liquid under
the necessary conditions of temperature and pressure.
A few days only after M. Pictet of Geneva had suc-
ceeded in liquefying oxygen, M. Cailletet of Chatillon-
sur-Seine liquefied not only oxygen and carbonic
oxide, but likewise hydrogen, nitrogen, and air.
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Gaseous and Liquid States. 5
Although there is thus no absolute distinction
between permanent gases and vapoiu^, it is convenient
to use the word * vapour ' to indicate a gas which
at ordinary temperatures can be reduced to the
liquid state.
Experiments by Dr. Andrews have shown that
gases in changing into liquids can be made to pass
through an intermediate condition, in which it is im-
possible to say to which of these two states of matter
they more exactiy correspond. By enclosing a vapour
in a tube, and subjecting it, at a very high tempera-
ture, to a considerable pressure, the vapour can be
made to pass by imperceptible degrees, i.e. without
any apparent optical change, into the liquid state.
For this purpose the vapour must be compressed to
that volume which it would occupy in the liquid state,
the temperature being sufficientiy raised to prevent
liquefaction from taking place. At a particular tem-
perature, which is known as the critical point, and is
diflferent for different gases, the tube is found to be
occupied by a homogeneous fluid which is neither
a liquid nor a gas, but which changes into one state
or the other by slightiy lowering or raising the tem-
perature, the volume remaining constant We thus
see that liquids and gases are convertible the one into
the other, and that matter can pass from one state to
the other without any perceptible gradations.
§ 5. Viscosity. — Fluids differ very widely with
respect to their distinguishing characteristic, viz. the
mobility of their particles. In a perfect fluid the par-
ticles are supposed to move among one another with-
out encountering any frictional resistance such as
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6 Hydrostatics,
retards the motion of one solid when moving on the
surface of another. But no fluid exists which fulfils
this condition. By agitating a fluid, that is, by causing
one part of it to move against another, heat is gener-
ated just in the same way as when two sticks of wood
are rubbed together. This shows that the particles
encounter frictional resistance to their motion.
Gases approach more nearly to the definition of a
perfect fluid than liquids. The latter are found to
exhibit every variety of difference with respect to the
mobility of their 4)articles, some approaching to a
semi-solid condition, and exhibiting properties inter-
mediate between the liquidity of water and the rigidity
of ice. Liquids such as treacle, new honey, and tar,
in which this frictional resistance appreciably interferes
with the mobility of the particles, are called viscous.
This viscosity, or * quasi- solidity,' as it is sometimes
called, is common to all liquids, and exists, though to
a small degree, in water. To this property is mainly
due the resistance which a vessel experiences in its
passage through the sea ; and it can be shown
that a body, completely submerged, and moving
with a uniform velocity through a perfect fluid, would
experience no resistance whatever to its motion.
§ 6. The Three States of Matter.— If we compare
together certain typical examples of solid, liquid, and
gaseous bodies, such as stone, water, and air, we find
that they exhibit distinct and characteristic properties,
by which they may be referred to different classes.
Keeping these differences in view, we have seen that
we can define a perfect fluid or a perfect liquid, al-
though we know very well that matter nowhere exists
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The Three States of Matter, 7
which fulfils the conditions of these definitions. But
when we consider the varieties of solid, liquid, and
gaseous bodies which experience brings under our
notice, we discover that matter actually exists in all
intermediate states between these three typical con-
ditions. Thus we have glue-like liquids of every kind,
between a clear limpid liquid and a gelatinous or
semi-solid mass ; and we have vapours in that critical
condition through which they can be made to pass
by imperceptible degrees from the gaseous to the
liquid state. No strict lines of demarcation can
therefore be drawn between these various conditions
of matter ; and the solid, liquid, and gaseous states
may be regarded as only widely separated forms in
which, under different conditions, the same substance
may exist
§ 7. Hydrodynamics defined. — The science which
treats of the application of the laws of Dynamics to
fluid bodies is generally known as Hydrodynamics.
The axioms, or fundamental principles, of the science
are Newton's laws of motion, which apply equally to all
branches of Dynamics. The mobility of the particles
of fluid bodies gives rise to important diflerences be-
tween their behaviour and that of solid bodies under
the action of external forces, which render convenient
the separate treatment of this subject.
Under the general head of Hydrodynamics are
included Hydrostatics and Pneumatics, or the study
of the laws of motion in their application to liquid
and gaseous bodies respectively.
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8 Hydrostatics,
II. Units of Measurement,
In the solution of problems in Hydrostatics a
knowledge of the weights of certain volumes of fluid
is very generally required. It is desirable, therefore,
to note at the outset the different standards of measure-
ment which are commonly employed.
§ 8. Units of length. — The imit of length gene-
rally used in England is one foot, which is one-third
of a yard. The imperial yard is an arbitrary measure-
ment, not derived from any fixed quantity in nature,
and is defined * as the distance between two marks on
a certain metallic bar preserved in the Tower of
London, when the whole has a temperature of 60° R'
In the French or metric system the stan^rd is
the metre, * defined originally as the ten-millionth part
of the length of the quadrant of the earth's meridian
from the pole to the equator ; but now defined prac-
tically by the accurate standard metres laid up in
various depositories in Europe.' The metre is some-
what longer than the yard, being equal to i '093623 11
yard, or to 39*370432 inches. The great convenience
of this system for ordinary purposes is the employ-
ment of decimal parts or multiples of the metre, to
represent smaller or larger units. Thus, in any ex-
pression, if the units represent metres, the tens re-
present rt^^f^^-metres, the hundreds hecto-mtixt^, and
so on. In the same way the first decimal place
represents deci-mttrts, the second cenfi-mtXxtSy the
third mi//i-mQtres, and so on. Thus 135724 metres
represents i hecto-metre, 3 deca-metres, 5 metres,
7 deci-metres, 2 centi-metres, and 4 milli-metres.
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U7tits of Length and Volume. 9
In physical investigations the centimetre is now
generally accepted as the unit of length. This unit has
been selected by a Committee appointed by the British
Association, and is recommended for general adoption
in a work * published by the Physical Society of Lon-
don. The system of units, based on the recommen-
dation of this committee, is known as the Centimetre-
Gramme-Second system of units, and is generally
referred to as the C.G.S. system.
I foot = 30*4797 cm.
When the number of units is very large, it is
expressed as the product of two factors, one of which
is a power of 10. Thus 3240000000 is written
3*24 X 10®, and 0*00000324 is >vritten 3*24 x io~®.
§ 9. Units of Area. — In England we use, com-
monly, the square yard, the square foot, and square
inch. In the metric system we have the sq. metre,
the sq. decimetre, sq. centimetre, &c.
I sq. metre =100 sq. decs. = 10,000 sq. centimetres.
In the C.G.S. system of units, the unit of area is
the square of the unit or length, i.e. i sq. centimetre.
I sq. foot =: 929*01 sq. cm.
§ 10. TTnits of Volume. — The advantages of the
metric system over that ordinarily used in this country
are most apparent when we have to compare units of
volume with units of length. In our system of mea-
sures no simple relation exists between these two units.
The gallon, which is the common unit of volume,
cannot be represented by any exact number of cubic
feet or inches. In the French metric system the unit
• * Illustrations of the C.G.S. System of Units,' by J. D.
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lO Hydrostatics.
of volume is the litre^ and a litre is equal to a cubic
decimetre. Thus : —
I cubic metre = looo litres.
I gallon = 4*54346 litres = 277*274 cubic inches.
In the C.G.S. system the unit of volume is the
cube of the unit of length, i.e. the cubic centimetre.
I cubic foot = 28316 cubic centimetres.
§ II. ITnits of Mass and Weight.— The British
unit of mass is the quantity of matter which weighs
one pound. It is defined by standard only.
The French standard is the kilogram, defined
originally as the quantity of matter in a litre of water
at 4° C, but now practically determined as the mass
of a particular piece of platinum preserved in the
Ministfere de ITntdrieure at Paris, and by standards
which have been compared with this.
In the C.G.S. system, the unit of mass is one gram,
and is equal to the mass of a unit-volume, i.e. a cubic
centimetre of water, at 4° C.
I grain = '0647990 grams.
As the weights of bodies are proportional to their
masses at places equally distant from the earth's
centre, the foregoing units of mass may be taken as
equivalent units of weight. Thus the weight of 3
cubic centimetres of water is 3 grams, and the volume
occupied by 3,000 grams of water is 3,000 cubic
centimetres.
§ 12. Unit of Time ; derived Units.— The unit of
time is one second. Units of velocity, acceleration,
momentum, force, energy, heat, &a, which are based
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Preliminary Notions. 1 1
on the fundamental units of length, mass, and time,
are called derived units. Thus in the C.G.S. system
the unit of force is that force which, acting upon a
gram for one second, generates a velocity of one
centimetre per second.
§ 13. Geometrical Eelations. — ^As problems fre-
quently occur which presuppose a knowledge of the
measurements of the areas and volumes of certain
figures, the following geometrical relations should be
remembered : —
(i) The ratio of the circumference of a circle to
its diameter = 3-14159 = fjg = \^ nearly,
and is represented by the Greek letter x.
(2) The circumference of a circle = 2 tt r, where
r is the radius of circle.
(3) The area of a circle = ir r*.
(4) The area of the surface of a sphere = 4 ir r*.
(5) The volume or contents of a sphere = | tt ^^.
(6) The area of the curved surface of a cylinder
equals the product of the height into the
circumference of the base = 2 tt rh,
(7) The volume of a cylinder equals the product
of the height into the area of the base,
= TT r'^,h,
(8) The area of the curved surface of a cone equals
the product of the slant side into half the
circumference of the base, = tt rs/h^-^r^y
where h is the height of the cone.
(9) The volume of a cone equals one-third of the
volume of a cylinder on the same base, and
of the same height, = 3^ t r'^,h.
Digitized by VjOOQIC
12 Hydrostatics.
III. Density, Specific Gravity,
§ 14. Density. — If we take two vessels of equal
capacity and fill the one loosely with some substance,
such as sand, and compress into the other a much
larger quantity of the same substancCy we should say
that the density of the matter in the one vessel was
less than that in the other. If, then, we understand
by mass quantity of matter, we see that the densities
of two bodies of equal volume and of the same material
are proportional to their masses.
§ 15. Measure of Density. — The measure of density
is the mass of a unit- volume. If we adopt the cubic
centimetre as the imit- volume, and the gram, or quan-
tity of matter in a cubic centimetre of water at 4° C.,
as the unit of mass, then the density of a body is
measured by the number of grams in a cubic centi-
metre of its substance ; and if //represent the density
of a body whose volume is V and mass M^ d is the
M
mass of a unit volume, and Jf = Vd\ or // = — .
§ 16. Specific Gravity. — If we take two vessels
of equal capacity, and fill the one with mercury and
the other with water, we shall find that the one con-
taining the mercury is heavier than the one filled with
water. This difference in the properties of the two
substances is known as a difference in their specific
gravities. When we say that lead is heavier than
wood, we mean that bulk for bulk the one substance
is heavier than the other — that a cubic foot of lead
weighs more than a cubic foot of wood.
§ 17. Measure of Specific Gravity.— The sp. gr.
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Density and Specific Gravity, 13
of a substance is measured by the weight of a unit-
vohime of that substance. If d be the mass of a unit-
volume, and s its weight, then s=:gd^ where ^ is the
acceleration due to gravity ; and if W^ be the weight
of a body whose volume is V and sp. ^. s, then
W=Vs,oxs=: ~
Since, also, s = gd, we have JV^gd, V,
If we adopt the weight of a gram as the unit of
weight, the specific gravity of a body is expressed by the
weight in grams of a cubic centimetre of its substance.
Ordinary definition of Specific Gravity. — The spe-
cific gravity of a substance is very often said to be
measured by the ratio of the weight of a given volume
of that substance to the weight of an equal volume
of some standard substance ; and in considering solid
and liquid bodies, water at 4° C. is taken as the stan-
dard ; whilst in the case of gases, air at 0° C. and 76 cm.
barometric pressure is employed. But it will be seen
that by defining specific gravity as the weight of a unit-
volume, we avoid the explicit reference to a ratio, whilst
the number expressing the ratio, when water is the
standard substance, is the same as the number of
grams representing the specific gravity. Thus, if
W be the weight of a given volume V of any sub-
stance, and W the weight of the same volume V of
water, then according to the ordinary definition
W -^ W '=' specific gravity. But if s be the weight
of a unit- volume of the substance, and a/ the weight of
a unit- volume of water, W^=^ Vs and W^ = Viv.
Vs s
Hence, specific gravity = -- - = — , and if wc take
Vw w
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14 Hydrostatics.
the weight of a unit- volume of water to be the unit
weight, as we have supposed, then the specific gravity
of a substance = j, the weight of a unit-volume of
that substance.
Since, also, the weight of a unit of mass is equal to
g units of force, and is represented in gravitation units
by one gram, we see that the numbers representing
the densities of bodies represent also their specific
gravities, and that the specific gravity of water is unity.
When the specific gravity is considered as a ratio, it
is sometimes called the relative specific gravity y to dis-
tinguish it from the absolute specific gravity ^ or weight
of the mass of a unit- volume.
Where, as in the English system of weights and
measures, the weight of a unit- volume of the standard
substance is not adopted as the unit of weight, the
specific gravity of any substance, considered as a
ratio, = J" -f- a/, where s is the weight of a unit- volume
of the substance, and w is the weight of a unit- volume
of the standard. If, therefore, S be the relative
specific gravity of the substance, we have ^5= Vs-^w
or VSw =zVs = lV.
Thus, if we wish to find the weight of 4 cubic
inches of zinc, we must first know the weight of a
cubic foot of water (a/), and then the specific gravity of
zinc (represented by a ratio) being 7, we have JF=
7 X ttVb- ^ ^y ^^d taking w to equal 1,000 oz. roughly,
we have H^= 7X4xiooo ^ ^.^^ 1^^^ ^^^^
16x1728
The weight of a cubic foot of water is more nearly
equal to 62*3 lbs.
In estimating the weights of gases, it is useful to
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Tables of Specific Gravity.
^S
remember that one cub. cm. of dry air at o° C. and
76 cm. pressure (at Paris) weighs 0*001293 grams.
Under the same circumstances, 13 cubic feet of air
weigh very nearly i lb. avoirdupois.
§ 18. SPECIFIC GRAVITY OF SOME IMPORTANT
SUBSTANCES.
TABLE I.
Solids.
Name of Substance
Specific
Gravity
Name of Substance
Specific
Gravity
Platinum, cast
Gold, cast .
Lead, cast .
Silver . . .
Bismuth . .
Copper, hammer^
„ wire
Brass . .
Nickel .
Steel . ,
Iron, wrought
Iron, cast
Tin . .
Zinc . .
Antimony
Iodine
Diamond .
Flint-glass
Aluminium
Bottle-glass
Plate-glass
Marble .
Emerald .
Rock-crystal
Porcelain
20-86
1925
11-35
10-47
9-82
8-88
878
8*39
8-28
7-82
779
7*21
7-29
7 'CO
671
4*95
3-52
378 to 3-2
2-57
2-6o
2-37
2-84
277
2-66
2*49 to 2*14
Sulphur, native
Ivory ....
Graphite . . .
Phosphorus . .
Magnesium . .
Amber . . .
Wax, white . .
Sodium , . .
Potassium . .
Ebony, American
Oak, English .
Mahogany, Spanish
Box, French
Beech . .
Ash . .
Maple
Cherry-tree
Walnut .
Pitch pine
Elm . .
Cedar
Willow .
Larch . .
Poplar
Cork . .
203
1*92
1*8 to 2*4
177
174
I -08
097
0-97
0-86
1*33
o'97 to I '17
I '06
1*03
0-85
0-84
075
071
0-68
0-66
060
059
0-58
0-54
038
0*24
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i6
Hydrostatics.
TABLE II.
Liquids, at o® C.
Name of Substance
Specific
Gravity
Name of Substance
Specific
Gravity
Mercury ....
Sulphuric acid . .
Nitric acid . . .
Aqua regia . . .
Hydrochloric acid .
Blood, human . .
Ale, average . .
Milk
Sea-water . . .
Vinegar ....
Tar
1-848
1*500
1*234
I -218
1*045
1*035
1-030
I '028
1-026
I 015
Water, distilled, at
Linseed oil . . .
Proof spirit . . .
OHve oil ... .
Ether, hydrochloric
Turpentine, oil of .
Brandy ....
Alcohol, absolute .
Ether, sulphuric .
I 000
0-940
0-930
0-915
0-874
0870
0-837
0796
0720
TABLE III.
Gases at 0° C. and 76 cm. Pressure.
Name of Substance
Specific
Gravity
Name of Substance
Specific
Gravity
Oxygen ....
Atmospheric air
Nitrogen ....
Hydrogen . . .
Chlorine ....
0001432
0001293
0001267
0000894
OX)03209
Hydrochloric acid
gas
Nitrous oxide . .
Carbonic acid . .
0-00164
0-00197
0-00198
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Density and Specific Gravity of Compounds, 17
§ 19. To find the density of a combination of
two or more substances whose volnmes and densities
are ^ven.
Let z/j, z/2, v^ ... be the volumes of the substances,
d\^ ^2> ^3«- their respective densities.
Then if Fbe the volume of the combination, and
if no contraction take place,
V=Vi +V2+ Vq 4- ...
and if Z) be the density of the whole, V£> equals the
mass of the whole, and therefore,
VD = Vi di +^^2 ^2 + ^'3 ^3 + •••
or Z) = i^lAjt" ^2 ^2 + z^3 ^8 + '"
Vi + >2 + v^ +...
If, however, as very frequently happens, contraction
takes place, and if the volume of the whole is some
proper fraction (= r) of the sum of the volumes of
the parts,
v,d, ^ v^d^^v^d^ + ...
^^^■^-r(z/i + z;, + z/3...)
A similar proposition holds good if, for density,
we substitute specific gravity.
§ 20. To find the specific gravity of a combina-
tion of substances, the weights and specific gravities
of which are given.
Let ze/i, 0/3, a/3 ... be the weights of the components,
and ^1, J2, s^ ... their respective specific gravities ; then,
if W be the weight of the whole and S its specific
gravity, we have IV = w^ + 0/3 + 0/3 4- ...
G
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iS Hydrostatics.
and since W = VS^ where V is the volume of the
whole
i^ — . ^ -4- -^ 4- J£a 4-
O Jj ^2 "^3
supposing no contraction to take place, and there-
fore
«$"! '$'2 "^3
But if the volume of the whole be less than the siim
of the volumes of the parts in the ratio of r : i, then
2— ^\ 4- «^2 + ^3 + ...
^\S^ ^2 ^3 ^
The density of a compound can be found in the
same way, the masses and densities of the components
being given.
§ 21. Examples.— (I.) i,ooo cc. of a gas whose density
is 12 are mixed witb 2,cxDO cc. of a gas whose density is i6,
and the volume of the mixture is diminished by one-third.
Find the density of the mixture.
T^ 1000x12 + 2000x16
U B _ = 22
f(l000 + 2000)
(2.) Three kilograms of a substance, sp. gr. = 4, are melted
with 5 kils. of a substance, sp. gr. = 6, and the volume of the
mixture is o*i less than the sum of the volumes of its com-
ponents. Find sp. gr. of mixture.
^ = 3 + 5 _ 8x240 _ -35
&(* + g) "" 9x38 '^^
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specific Gravity of Compounds, 19
Exercises. I,
1. Find the weight of a bar of copper, the section of which
is 64 square centimetres, and length one metre. ^
2. Find the size of an iron shot that weighs 60 lbs. ^/ v, c^c -
3. Find the weight of a column of mercury 30 inches high,
having a uniform section of one square inch. 1 1^ r*
4. A cubic inch of a substance weighs 2 oz. ; find its specific
gravity. J. ^
5. If silver and copper are mixed in the proportion of 2 : 7
by weight, find the specific gravity of the compound f y Q
6. What mass of copper must be mixed with 200 grams of
gold to make the specific gravity of the compound 0*9 of that
of gold? >.:^,^^W.-^ 2,/. /)3*w-
7. Equal weights of two fluids, of which the specific gravities
are s and 2j, are mixed together, and the mixture occupies three-
fourths of the sum of the volumes of its components. Find the
specific gravity of the mixture. ^)^
8. Three litres of a gas, the density of which is unity, are
mixed with one litre of a gas density 14, and the mixture occu-
pies half the volume of the original gases. . Find its density. ?i-^
9. A bar of cast-iron, sp. gr. = 7*2, is found to weigh 17-5
kils. The length of the bar is 5 decimetres, and sectional area
50 sq. centimetres. Is there a flaw in the casting ? If so, what
Is its size? l^Qj^ ^^^
C3
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20 Hydrostatics.
CHAPTER 11.
FLUID-PRESSURE ON SURFACES IMMERSED.
IV. Explanation of Terms. PascoTs Principle,
§ 22. Fluid Pressure. — The base and sides of a
vessel containing a liquid in equilibrium are subjected
to a certain pressure, which is caused by the weight
of the liquid, and by the resistance which the surface
of the vessel offers to the free motion of the liquid.
§ 23. Intensity of Pressnre.—By intensity of
pressure at a point is meant the pressure on the unit
of area containing that point
If a small surface a^ in contact with a fluid, is
maintained in equilibrium by a force jP, acting perpen-
p
dicularly to the surface, then - measures the average
a
pressure-intensity at any point of the area ; and if
P
-= /, then/ is the pressure on a unit of area.
a
§ 24. Variation of Pressure with DeptL— Sup-
pose the small horizontal area a to be at a depth i
below the surface of the liquid, then the area a will be
pressed vertically downwards by a force equal to the
weight of the column of liquid above it ; and if x equal
the weight of a unit volume of liquid, this force is equal
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Variation of Pressure with Depth, 21
io z as. If, then, / be the pressure at any point of
this area, p .a^= zas, orp = z s.
If a is not horizontal, the pressure at some parts
of the area is greater than at others. But if we sup-
pose this area to become smaller and smaller, and
ultimately to diminish without limit, the difference of
pressure at different parts of the area will diminish
likewise ; and if / represent the pressure intensity,
when the area has become so small that it may be
regarded as a point, then/. a^=z.a. s, where a is the
infinitely small area about this point, and z is its
depth. Hence/ = z,s, as before, showing that /;/ t/ie
same liquid t/ie pressure varies with the depth.
This proposition may be experimentally verified by
taking a small cylindrical vessel a (fig. 2), open at
both ends, and having a movable base o, to which a
thread c is attached. If we connect the base with one
end of a scale-beam and press y\c, a.
it against the vessel by at-
taching a weight W to the
other end of the balance, we
shall find, on pouring water
into the vessel, that the disc
falls off when the water has
risen to a certain height, and
that this height is always pro-
portional to the excess of the
weight W above the weight '
of the disc. This shows that
the ratio of the pressure on the disc to its depth below
the surface of the liquid is constant, />. that the pres-
sure varies with the depth.
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22 Hydrostatics,
By fixing the vessel a into another vessel contain-
ing water, it may be shown that the liquid in the outer
vessel exerts an upward pressure on the under side
of the disc equal to the downward pressure of the
liquid contained in the inner vessel. For, if we attach
the disc o to one end of the scale-beam, and to
the other end a weight sufficient to counterbalance it,
we shall find, on pouring water into a, that the disc
does not fall off until the level of the water in the two
vessels is the same,
§ 25. Direction of Pressure on a Surface in Con-
tact. — If a thin plate be immersed in a fluid and
be held in any position, the direction ot the pressure
is perpendicular to the surface of the plate, if the
fluid is at rest. For if it acted in any other direction
it could be resolved into two components, one per-
pendicular to the surface and the other along it;
and the latter component would, in the absence of
friction between the surface and the fluid, produce
motion, which is contrary to supposition. The direction
of the pressure must, therefore, be perpendicular to
the surface.
§ 26. Equal Transmissibility of Fluid Pressure.
If we take a vessel full of water (fig. 3), having
various apertures of the same size, fitted with water-
tight pistons which are kept in equilibrium, and
if one of these a be pressed downwards with a force
p, an additional pressure equal to p will be required
at each of the other pistons to preserve equilibrium.
Thus if a force of i lb. be applied at a, it will be
necessary to apply an increased pressure of i lb. to
each of the other pistons to prevent motion ; and as
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Equal Transmissibility of Fluid-pressure, ^3
it matters not in what part of the vessel these pistons
Fig.
are fitted, we see that any
increase of pressure applied
at one part is transmitted
equally throughout the fluid.
If the pistons, as in fig. 4,
are of different areas, that
of B being twice that of a
and the area of c being
nine times as great, it will
require an additional force
of 2 lbs. at B, and of 9 lbs.
at c, to preserve equilibrium, when a force of i lb.
IS applied at a.
These experiments serve to
illustrate the following fundamental
law of fluid-pressure, which is
known as Pascal's ^ principle :
When pressure is communicated
to any part of a fluids it is trans-
mitted equally in all directions
through the fluid.
§ 27. Hechanical Application.
principle, which is a ne-
necessary consequence of
the mobility of the par-
ticles of a fluid, serves to
explain the action of a
very useful mechanical
contrivance for multiply-
ing power.
* Pascal was bom at Auvergne, 1623 ; dkd 1662.
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24 Hydrostatics.
It consists of two communicating vessels contain-
ing water, one of which is much larger than the
other. The vessels are fitted with pistons P and
/, the areas of which we will suppose to be A and
a. If now weights ?f^ and zc' be placed on those two
pistons respectively, so as to counterbalance one
another, it will be found that W \w\\ A : a, which
is in accordance with Pascal's principle.
We see, also, that if the piston / be pressed down
through the space S, the water contained in the
smaller vessel will pass into the larger, and force up
the piston F through some space s, such that—
ax SssAxs,
since the volume of water that is removed from one
vessel is the same as that which enters the other
vessel. Hence
S^A_W
saw
or, w, S=i W, s,
/>., the work done by 7V = work done by W.
§ 28. Hydrostatic Paradox. — A consequence of
Pascal's principle is that a quantity of water, however
small, can be made to support a weight, however large,
and this seeming paradox can be exhibited in the
following manner.
Let A B (fig. 6) be a long narrow pipe communi-
cating with a vessel c d, into which a piston c e is
fitted. If now water be poured through the pipe, it
will be found to rise to the same level, c e f, in
both parts of the vessel ; but if a weight Wht placed
on c E, and additional water be poured into the pipe,
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Flow dependent on difference of Surf ace4eveL 25
a position of equilibrium will result such as is shown
in the figure. In this case, the pressure exerted by
the water in the pipe above
the plane c e f supports the
weight W. Since this pressure
is communicated to every
element of area of c e equal
to the sectional area of the
pipe, the whole pressure on
c E is as many times greater
than the weight of the water
in A F as the area of c e is
greater than the area of the
pipe.
Let a equal the sectional area of the pipe ; then if
P is the pressure at f produced by the weight of water
above it, /^ = dJ x a f x a/, where w is tl\e weight of
unit-volume of water, and if A is area of piston c E,
the pressure communicated to c e is
— XaXAF XW= AXAFXW
a
/. W =iA X A F X a/.
This shows that the weight IV can be made as great
as we please, by taking A f or c e sufficiently great,
and that it is independent of «, the section of the tube.
Hence, with a tube sufficiently narrow, a quantity of
liquid, however small, can, in principle, support a
weight, however large.
§ 29. Communicating Vessels containing Liquid.
If two vessels a and b (fig. 7) containing the same
liquid be connected by a pipe, and if the surface level
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26 Hydrostatics,
of the liquid in a be higher than that in B, the liquid
will be found to flow from a to b.
This is easily explained by considering the pres-
sure on either side of any sec-
tional area, / ^, of the pipe.
For it is evident that the pres-
sure on the side towards A is
greater than on the side to-
wards B, since the pressure
varies with the depth below
the free surface, and consequently the flow takes place
from A to B.
It should be observed that although in the case
now considered the liquid flows from a to b, the
pressure-intensity along the base of a is less than
that along the base of b , and also that the pressure
at a^ where the fluid escapes from the one vessel, is less
than that at ^, where it enters the other vessel.
If we connect by a tube c d two parts of the
same vessel, although the pressure at d is greater than
that at c, no flow takes place, for it will be seen that
the pressure on either side of any element of area in
the pipe c d is the same.
We see, therefore, that a liquid always flows from
places of higher to places of lower surface level, and
that if a liquid is in equilibrium its surface level must
be uniform. Hence it follows that if a liquid be con-
tained in a vessel, its surface must be horizontal ; or,
more generally, every point of the surface of a liquid
in equilibrium must be at the same distance from the
earth's centre. In the case of large inland seas, the
surface is somewhat curved.
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Communicating Vessels, t.'j
We also see that the pressure at all points in the
same horizontal plane within a liquid at rest must be
the same, since these points are all equally distant
from the free surface, and the pressure varies with the
depth (§ 24).
It also follows that if any number of vessels con-
taining liquid communicate with one another, the
liquid will stand at the same level in all. For if the
surface level were higher in any one vessel, a flow of
liquid would take place from that vessel into the
others, till the uniformity of surface-level had been
established.^
This fact may be experimentally verified by pour-
ing water into one of the parts fig. 8.
of a vessel similar to that ^^^ M, £/ £f
shown in the figure, when it ^^ ^H S M
will be found that the free sur- |^^^^^^^^gi
faces of the liquid in all parts ^
of the vessel lie in the same horizontal plane. The
tendency of liquids to find their own level, />.
to flow from places of higher to places of lower
surface-level, is of great practical importance. It
is utilised in the water-supply of towns. A reser-
voir of water is kept at a considerable elevation, and
from it pipes proceed in all directions conveying water
to any heights below the level in which it stands in
the main reservoir. The water-level is an instru-
ment which acts on the same principle. It consists
of a tube bent at right angles, and furnished at its
^ It should be observed that difference of surface-level cor-
responds with difference of temperature in heat, and with
difference of potential in electricity.
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28 Hydrostatics.
extremities with two glass vessels, and the whole is
partly filled with water. Since the liquid has the
same surface-level in both branches, any points which
the observer's eye detects to be on the same level
with the surface of the water in both branches, will
be in the same horizontal plane.
When a liquid falls from a higher to a lower level,
there is a change of potential into kinetic energy,
and a corresponding amount of work can be effected
This fact is economically employed in water mills.
§ 30. Examples. — (i.) Find the whole pressure exerted on
a horizontal area of 9 sq. cms. which is sunk 125 cms. below
the surface of water. The pressure equals the weight of water
supported,
= 125x9=1125 grams.
(2.) Find the pressure- intensity due to a column of 25 cms.
of mercury upon which rests a column of 50 cms. of water.
The pressure-intensity is the pressure per unit area, i.e, per
sq. cm. :
= 25 X 13 '6 + 50 = 390 grams = 390 g units of force,
the specific gravity of mercury being 13*6.
(3.) Find the vertical force necessary to support the hori-
zontal base of a vessel containing mercury, if the area of the
base is one square decim., and its depth below the surface of the
mercury 5 centimetres, neglecting the weight of the base.
°=^ d! j« 5 X 100 X 13*6 grams.
.•. P^ 6800 grams.
(4.) What must be the height of a column of mercury to
exert a pressure of 1220 grams per sq. centim. ?
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Fluid-pressure on Surfaces immersed, 29
If A be the height required, I220s^ x 13*6 grams.
. , 1220 o_^
. , h = — - = 897 cm
136
§ 31. Equilibrium of two different Liquids in
Conmnmicating Tubes. — If two liquids that do not
mix meet in a bent tube, or in two tubes communicat-
ing with each other, the heights of their free surfaces
above their common surface are inversely proportional
to their specific gravities.
Let A a' be a plane drawn through the common
surface of the two liquids in one of the tubes. Let s and
^ be their respective specific gravities. Let ff and c be
the free surfaces of the liquids. Then p^^
if the liquids are in equilibrium, the
pressures at a and a' must be the same.
The pressure-intensity at a is « ^ x jt,
where ab \% the vertical height of b
above a.
The pressure-intensity at a' is « ^ x j'
^vhere acvst vertical height of c above a'.
Hence ab y. s ^ ac x s'
or ab : ac W s' : s.
Exercises. II.
1. Two communicating vessels contain fluid, and are fitted
with pbtons, the diameters of which are 2 inches and 8 inphes
respectively. If a weight of 3lbs. is placed on the smaller
piston, what weight must be placed on the larger to preserve
equilibrium ? ' '
2. A narrow vertical pipe is attached to a vessel (fig. 6),
Which is fitted with a piston the area of which is 2 square deci-
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30 Hydrostatics,
metres. If the vessel and pipe contain water, find the height
of the water in the pipe when a weight of 40 kils. is placed on
the piston, '^ \ ^
3. A cylindrical vessel contains mercury to the height of 2
inches above the base, and a layer of 8 inches of water resting
on the mercury. Find the pressure at any point in the base,
taking sp. gr. of mercury to be 13 '6.
4. At what depth below the surface of a lake is the pressure
intensity 5 times as great as at a depth of 10 feet, supposing
the atmospheric pressure to be equal to the weight of a column
of water 34 feet high ?
5. Two liquids that do not mix are contained in a bent
lube ; the difference of their levels is 3 ins., and the height of
the densenabove their conmion surface is 5 ins. ; compare their
specific gravities.
6. If, in the above, the internal section of the tube is one
square inch, and the lighter liquid is water ; find the weight of
water contained in the tube.
V. Whole Pressure on Surface immersed,
§ 32. Whole Pressure. — When a vessel contains
a fluid, or when a body is immersed in a fluid, the
fluid exerts a normal pressure at each point of thesur-
FlG. 10.
face in contact with it. The sum of all these pressures is
called the whole pressure on the surface immersed.
It should be observed that these pressures act in
(Jifferent directions, the pressure at each point being
perpendicular to the surface at that point The whole
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WItole pressure on a Surface immersed. 31
pressure is the sum of all these pressures, and repre-
sents the total strain to which the vessel containing
the fluid, or the body immersed, is exposed.
§ 33. To find the Whole Pressure which a
Liquid exerts on a Surface immersed. — Let a b c d
be any area in contact with a fig. 11.
homogeneous liquid, the free
surface of which is the horizon-
tal plane abed. Let ef be
any element of area so small
that the pressure-intensity at
all points may be supposed
to be constant, and let the
depth of this area below the
surface of the liquid be z. Then if s be the weight of
a unit- volume of the liquid, the pressure exerted one f
\% a z s^ where a is the area of ^/
Now the whole pressure is equal to the sum of the
normal pressures on all the elements that make up the
whole area. Therefore the whole pressure is equal to
ai Zi s + a2 Z2 s -^ a^ z^ s + . . .
= {^1 Zi -\- a^ Z2 + a^ z^ -\r . . .) s.
But, by the properties of the centre of gravity,
a^ Zi + ^2 ^2 + ^3 ^3 + • • • = (^1 + ^2 + ^3 + • • •) ^
where z is the depth of the centre of gravity of the
whole area below the plane surface of the liquid ;
. '. whole pressure ^= A z s,
where A is the whole area in contact, z the depth of its
centre of gravity below the surface of the fluid, and s
the weight of a unitvolume of the liquid. Or, the
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32 Hydrostatics.
pressures Azdg where// is density of fluid and g the
acceleration due to gravity. Hence, The whole pres-
sure on any area immersed equals the weight of a column
of liquid which has that area for base, and the depth
of its centre of gravity below the surface of the liquid for
height.
This proposition holds good whatever may be the
shape of the vessel containing the liquid ; and what-
ever may be the position of the area immersed. In
some cases the results to which it leads seem at first
sight paradoxical, but they will be shown to agree
with the general proposition which we have now esta-
blished.
§ 34. Examples. — (i.) Compare the pressures on the base
aiid side of a cube filled with liquid.
Let the edge of the cube be a, then area of the base is a\
and depth of centre of gravity below surface is £Z,
/. whole pressure on base is a* x a x s^c^s.
The area of a side is also a^ ; but depth of its centre of gravity is
~, /, pressure on side istf^x — x j«— j,
2 22
,'. whole pressure on base =» twice the whole pressure of one
of the sides.
The pressure on the base acts vertically, and the pressure on
the sides horizontally.
(2.) An oblong, the edges of which are 6 metres and 8 metres,
is immersed vertically with its shorter edge horizontal, and 2
metres below the surface of water ; find the whole pressure on
either side.
Area immersed is 6 x 8 = 48 square metres.
Depth of centre of gravity is 4 + 2 = 6 metres.
/, whole pressure is 48 x 6 x j, where s is the Weight of
a cubic metre of water, i,e, 1,000 kilograms.
/. Pressure - 288,000 kils.
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Fluid-pressure on Surfaces immersed. 33
Fig.
(3.) A right pyramid, the height of which is 8 decs., has a
square base, each edge of which is 5 decs. Required the whole
pressure on the base when the pyramid is full of liquid. The
area of the base is 25 sq. decs., and the
depth of its centre of gravity below the
sur&ce of the liquid is 8 decs. Hence
the whole pressure on the base is 80
X250OXJS20OX kils., where s is the
sp. gr. of the liquid. It is to be observed
that the pressure on the base, in this
case, is greater than the weight of the
liquid the vessel contains, being equal
to the weight of liquid in the prism hav-
ing the same base, ue, to a column of
liquid the height of which is E F, and base A B c D. On the
other hand, the pressure transmitted to the stand on which the
vessel rests is equal only to the weight of the liquid contained
in the vessel ; and consequently the pressure on the base of the
vessel is greater than the pressure communicated to the stand.
This result which follows directly from the general proposition,
requires further explanation, and will be considered in the fol-
io viring paragraph.
§ 35. To find the whole pressure exerted by a
liquid on the base and sides of the vessel con-
taining it. — ^We have three cases to consider, ac-
cording as the sides of the vessel are vertical or
slant from the base outwards or inwards. We shall
suppose the base of the vessel to be horizontal If
the sides are vertical, as in fig. 13, the pressure on the
base is evidently equal to the weight of the fluid con-
tained in the vessel, and the horizontal pressures are
equal
Suppose now that the sides are inclined, outwards
in fig. 14 and inwards in fig. 15.
The pressure at any point o in the side a b is a
Digitized by VjOOQIC,
.^4
Hydrostatics,
force p acting perpendicularly to a b. As this force
is prevented by- the resistance of the slant side from
Fia
15.
E
-.^-U
r
r
-^*
^--j
causing motion, it produces a reaction equal in mag-
nitude but opposite in direction. This reaction /
can be resolved into two components, x and y acting
at o, X horizontally in both figs., and y vertically up-
wards in fig. 14 and I vertically downwards in fig. 15.
It thus appears that in fig. 14 part of the weight
of the liquid is supported by the sum of the forces
y acting at all points in the slant side, and the
remainder of the weight of the liquid presses on the
base B c. Hence, in this case the pressure on the
base is /ess than the weight of the contained liquid.
But in fig. 15 the force y acts downwards, and
consequently increases the pressure on the base caused
by the weight of the liquid ; and hence the pressure on
the base of the vessel is greater than the weight of the
contained liquid by the sum of the vertical components
of the reactions caused by the pressure of the liquid
against all points in the slant sides.
Again, since the force /, due to the pressure of
the liquid, and the components, x and y of the reac-
tion of the surface a b, are in equilibrium, they can
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Pascal's Vases. 35
be represented by the sides of the triangle a e b. In
the same way the forces/', ^',y, acting at q can be
represented by the sides of the triangle d f c
Hence/ :^:;c::ab:ae:eb and
Z' : J'' : ^ : : c D : D F : F c
But E B is equal to f c, when the base is horizontal
.-. X = x' \ and in the same way the horizontal
pressures at all other points can be proved to be
equal. Hence the horizontal components of the pres-
sure of the liquid against the slant side are equal
If the base is not horizontal, then the whole pres-
sure on the base can be resolved into vertical and
horizontal components, and the algebraic sum of the
horizontal pressures on the base and sides of the
vessel will still be found to equal zero.
§ 36. Experimental verification. Pascal's Vases.
— In order to verify the general proposition, which we
have now proved, viz., that the pressure on the base of
a vessel containing liquid is independent of the shape
of the vessel, and varies only with the area of the base
and the depth of its centre of gravity below the sur-
face of the liquid, Pascal contrived an experiment very
similar to the following :
Take three vessels . p, q, m, of different shapes,
having the same circular aperture at their base,
and capable of being screwed into the ring of the
stand A B. One of the vessels being fastened to the
ring, a circular disc d hanging from one arm of the
balance is pressed against it by weights placed in the
scale-pan hung to the other arm. Water is then
poured into the vessel, and an index / marks the
D 2
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36 Hydrostatics.
level at which the water stands when the disc falls
off If the experiment is now tried with the other
Fig. x6.
vessels, it is found that the disc falls off when the water
rises to the same level. This shows that the pressure
on the base of each vessel is the same when the water
is at the same height above it
Exercises. III.
1. Find the whole pressure on a rectangular surface 6 feet
by 4 feet, immersed vertically in water with the shorter side
parallel to, and 2 feet below the surface.
2. Find the whole pressure on the curved surface of a
vertical cylinder which is filled with a liquid, sp. gr. = I '5, the
height of the cylinder being 2 decims., and the radius of the base
7 cms.
3. Show that if a sphere or a cube be filled with liquid the
total strain to which it is subjected is three times the weight of
the liquid it contains.
4. A flood-gate is 6 feet wide and 12 feet deep. What is
the total pressure on the flood-gate when the water is level witir
the top?
5. A globe, tha radius of which is 3*5 cms., rests at tho
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Centre of Pressure. 3 7
bottom of a vessel 30 cms. in height, foil of water. Find the
total pressure on the globe.
6. A conical vessel 2 decimetres high, has a movable base,
of 25 sq. cms. area, formed by a disc 2 cms. thick, and the
specific gravity of the material being 3 '2, find the force neces-
sary to uphold the disc when the vessel is full of water.
7. A smooth vertical cylinder, 2 feet high and i foot in
diameter, is filled with water, and closed by a piston weighing
3 lbs. Find the total pressure on the curved surface.
VI. Centre of Pressure.
§ 37. Definition. — When a plane surface is im-
mersed in a fluid, the pressures at different points of
the surface are perpendicular to it (§ 25), and consti-
tute a system of parallel forces of which the whole
pressure is the resultant. The magnitude of this re-
sultant pressure we have seen how to find, but the
point at which it *acts we have not yet determined.
This point is called the centre of pressure^ and may be
defined as the point of action of the single force equiva-
lent to the whole pressure exerted by a fluid on any plane
surface with which it is in contccct
If a plane surface is immersed horizontally the
centre of pressure corresponds with the centre of gravity,
but not so if it be immersed in any other position.
For in this case the pressures on equal elements of
area are not equal, since the pressure varies with
the depth (§ 24), and the different elements of area
into which the whole surface may be divided are
at different depths below the surface of the fluid.
Consequently, the centre of pressure, which is the
point in, the plane surface at which the resultant of
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3»
Hydrostatics.
all these forces acts, does not correspond with the
centre of gravity of the surface.
The term centre of pressure is used with respect
to plane surfaces only, since it is not always possible
to find a single force equivalent to the resultant action
of a fluid on a curved surface.
§ 38. To find the centre of pressure of a rectan-
gpilar area immersed vertically.
Let A B c D be a rectangle having its upper side
A B in the surface of the
liquid Then ifEF be drawn
to bisect A B and d c, the
pressure will be equally dis-
tributed on each side of e f,
and the centre of pressure
will lie somewhere in this
line.
To •determine where,
take M N bisected by f to
represent the pressure at r. Join em, en. Then if
we take any point p in e f, and draw q p r parallel
to M F N, Q R will represent the pressure at p.
For, since the pressure varies with the depth, the
pressure at p : the pressure at f : : e p : e f
and ep:ef::qr:mn
/. the pressure at p : the pressure at f : ; q r : m n
and /. Q r represents the pressure at p.
Now the problem of finding the point of action of the
resultant of a number of forces represented by such
lines as q r, acting at all points of e f, is the same as
that of finding the centre of gravity of the triangle e m n.
But the centre o^ gravity of this triangle is kno\ni to be
I> M
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Centre of Pressure, 39
at a point g in e f, such that f g = J f e, and therefore
the centre of pressure is at the same point, i,e. at a
distance of one-third up the middle line from the
base.
§ 39. To find the centre of pressure of a triangle
having one of its sides in the surface of the liquid.
If the triangle be divided into a number of nar-
row horizontal stnps, the pressure on each strip acts
at its middle point, and therefore the whole pressure
on the triangle acts somewhere in the median line
drawn through the middle point of the side in the
surface of the liquid. But the pressure on each strip
is proportional to its area multiplied by its depth,
and the value of this product is constant for every
pair of strips at equal distances from the middle
point of the median line. Hence the resultant
pressure acts at the middle point of this line, or the
depth of the centre of pressure is half that of the apex
immersed.
§ 40. To find a general expression for the centre
of pressure of a plane surface.
Suppose the surface divided by horizontal lines
into any number of small elements, then if a be the
area of one of these elements and z its depth (which
may be considered the same as the depth of its
centre of gravity) below the surface of a liquid of
sp. gr. J, azs is the pressure on that element of
area.
Hence, if -s^i, 2^2, . . . . be the depths of the several
elements, the whole pressure equals
a\Z\S-\'aiiZi^s-\-a^z^sAr
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40
Hydrostatics.
and if Z be the depth of the centre of pressure, then
by the principle of parallel forces
or Z= ^i^i^+^2^^+g3V+. . . . .
The general application of this method, as of the
similar method for finding the centre of gravity of a
surface, requires the use of the integral calculus, but
special cases may be determined by ordinary alge-
braical processes.
§ 41. Bzamples. — (i.) To apply the method of § 40 to
find the centre of pressure of an isosceles triangle with its
apex in the surface of the liquid and its base horizontal.
Suppose the triangle to be divided by horizontal lines into
narrow horizontal strips.
Let h — height of triangle, h its base ; and let s be the
sp. gr. of the liquid.
Let h be divided into n
parts, then the breadth of each
of these strips will be -.
Also,
since F E ; E A : : B D
: DA
/. F E = — A E, and the area
2h
of each strip may be represented
by - • A E •— - = — A E, and
2h ft 2n
therefore the pressure on each strip may be taken as equal to
- / A E y. ^, supposing the depth of the strip below A to
correspond with that of its centre of gravity. Now since
— , we have
n n n n
A E has the several values — , ?-» 3 f .
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Fluid-pressure 09t Surfaces immersed. 41
z -
±
2n
n
h_
n
h_
4
h_
4
I^ + 2» + 3^+
I«+ 22+32 +
. 4
236
236
2« 3 6»*
(by summation of series.)
and if n be infinite
each of the quantities — and \ has for its limit zero,
ft fr
Hence Z ^ ^ , h.
4
Or, the centre of pressure is on the median line, and at a depth
of three - quarters of the
height of the triangle, from
its apex.
(2.) Required the mag-
nitude and position of the
resultant pressure on a flood-
gate, the level of the water
being different on either side.
Let A B be a section of
the flood-gate, and let the
height of the water on one
side be a, and on the other
side ^, and suppose h greater
than a.
Digitized by VjOOQ IC
Fia
19.
^
■^SZ—
rzz-:^.
e;:::^
^.
IT"
z:r:zr:irz:=~-
Birz:
7^:zz
H
?-s^^
^
:^
-J
j"^
/^2 Hydrostatics,
Then, if >6 be the width of the gate, the total pressure on
one side will be —
2 2
^p.
and on the other side — ^^5= Q ; and P acts at a point c such
2
that B c=s one-third of Bw, and Q acts at a point D such that
B D =: one- third B« (§ 38). The resultant of these forces Q-P
equals (^— a') — , and the point E where it acts can be deter-
mined by the principle of parallel forces. Thus :
* ' 2 2 2
2\ 3 3/
• B ^-t±hist
Exercises. IV.
1. A cubical block each edge of which is 5 cm. is sunk
in water, with two opposite faces horizontal ; find the difference
in the pressures on its lower and upper surfeces.
2. A cylinder 10 inches high contains liquid to the height
of 8 inches : find the line of action of the total pressure on the
interior surface of the cylinder.
3. A hollow cube is three-fourths filled with water. One of
the sides of the cube moves freely about a hinge at the base.
Required the force that must be applied at the upper edge of the
moveable side and perpendicular to it to keep it in equilibrium.
4. Find the height of a cylinder the diameter of which is
2 feet, so that the whole pressure on the curved surface may
be four times as great as the pressure on the base, when the
cylinder is filled with liquid.
5. A rectangle is immersed in water with one side in the
surface. Show how to divide it by a horizontal line into two
parts on each of which the whole pressure shall be the same.
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Fluid-pressure on Surfaces immersed, 43
6. A rectangular v«sel has a partition 10 inches high and
8 inches broad. On one side of the partition is water to the
height of 4 inches : on the other side alcohol (sp. gr. o*8) to the
height of 6 inches. Find the magnitude and position of the
resultant pressure on the partition.
7. An equilateral triangular lamina is immersed in water
with one side in the surface and the opposite angle 2 deci-
metres below its surface. If each side of the triangle mea-
sures 6 decimetres, find the total pressure which the water
exerts on it.
8. A rectangle a BCD is immersed in water with the
side A B in the surface. Find the pressure on each of the tri-
angles formed by the diagonal a c, and show where it acts.
Show also that the resultant of these two pressures coincides
with the pressure on the whole rectangle.
9. Find the height to which water may rise on one side of
a wall 2\ metres high and half a metre thick without overthrow-
ing it, the specific gravity of the material of the wall being 2.
10. A cylinder, height h inches and diameter 2r, contains
three liquids of specific gravities .S",, S^ .S",, which do not mix
in layers of equal thickness : find the whole pressure on the
base.
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44 Hydrostatics.
CHAPTER HI.
FLUID PRESSURE ON BODIES IMMERSED.
VII. Resultant Vertical Pressure, — Principle of
Archimedes,
§ 42. We have seen (§ 32) that if a body is
wholly immersed in a fluid, every point of its surface
is subjected to a pressure perpendicular to the surface
at that point.
Now, all these pressures, acting as they do in
various directions, can be resolved into horizontal and
vertical components ; and since the horizontal pres-
sures equilibrate each other, the resultant pressure
must be vertical, and act upwards or downwards.
Moreover, since the pressure varies with the depth, it
is clear, speaking roughly, that the whole pressure on
the lower half of a body is greater than that on the
upper half; and, hence, the resultant of all the
pressures on a body immersed is a force acting verti-
cally upwards. This force is called the resultant
vertical pressure,
§ 43. Experiment. — Take an ordinary weight of
one pound, and having attached it to a piece of strong
thread, let it hang wholly immersed in water from one
of the scale-pans of the hydrostatic balance, as shown
in fig. 22. It will now be found to weigh less than
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Fluid-pressure on Bodies immersed, 45
I lb., and this difference of weight is due to the re-
sultant vertical pressure acting upwards.
§ 44. Measure of the Besultant Vertical Pres-
sure. — If a symmetrical body with vertical sides be
immersed in a fluid, we can easily see what the mag-
nitude of this resultant pressure is. For, since the
downward pressure exerted by the fluid on the body
is equal to the weight of the column of fluid having a b
(fig. 20) for a base, whilst the upward pressure is equal
to the weight of the column having c d for a base, the
resultant vertical pressure must be equal to the dif-
FiG. 20. Fig. 21.
ference between the weight of these two columns, i,e,
to the weight of a column of fluid equal to a b c d, i,e,
to the weight of the fluid displaced.
If the body be of irregular shape, a more general
method of proof must be employed. Thus :
Suppose the body immersed to be a portion of the
fluid itself solidified. Then, if no change of density
take place, the solidified fluid will remain as before in
equilibrium. Hence the weight of this portion of the
fluid, which acts, at its centre of gravity, vertically
downwards, must be counterbalanced by the upward
pressure of the fluid ; and consequently the resultant
Digitized by VjOOQ IC
46 Hydrostatics.
vertical pressure equals the weight of the solidified
fluid. But the fluid would exert exactly the same
pressure on any other body occupying the same space
in the fluid. Hence the resultant vertical pressure on
any body immersed is equal to the weight of the fluid
displaced, and acts at its centre of gravity, which point
is called the centre of displacement^ or centre of
buoyancy. The principle thus established is commonly
known as the principle of Archimedes ^"^ and may be thus
enunciated :
When a body is immersed in a fluid it is subject
to a force equal to the weight of the fluid displcued^
which acts at the centre of buoyancy vertically upwards.
Or, A body immersed in a fluid loses a portion of its
weight equal to the weight of the fluid displaced.
Before proceeding to consider some of the chief
deductions from this proposition, we will show how
it may be experimentally verified.
§ 45. Experiments. — i. Take a vessel witii a spout
in one side, as shown in fig. 22. Pour in water till it
Fig. 22. begins to run out from the spout.
Take a piece of iron weighing i lb.,
and having suspended it by a fine
thread from one of the scale-pans
of the hydrostatic balance, weigh it
in the water, allowing the water
displaced to escape into another
vessel, B. The piece of iron will
^ be found to weigh nearly 14 oz.,
and the weight of the water col-
lected in the vessel b will be found to be a little
* Born at Syracuse in Sicily, flourished about 250 B.C.
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Principle of A rchimedes, 47
more than 2 oz., thus showing that the loss of
weight of the body is equal to the weight of the water
displaced.
2. Take a hollow brass cylinder, a, into which
a solid cylinder, b, exactly fits. Hang the hollow
cylinder to one of the Fig. 23.
scale-pans of the hy-
drostatic balance, and
attach the solid cylinder
to it by means of the
hook, as shown in fig. 23.
Now weigh carefully the
two cylinders. Having
observed their weight,
take a vessel containing
water, and let the solid
cylinder hang in it completely immersed. Equi-
librium is destroyed, and the free scale-pan de-
scends. Now fill the hollow cylinder with water,
and equilibrium is at once restored, clearly showing
that the weight of the water in the hollow cylinder,
i,e, the weight of a quantity of water of the same
size as the body immersed, is equal to the loss of
weight of the solid in water, Le, to the upward pressure
which the liquid exerts on the body.
§ 46. Beal and apparent Weight of Bodies. —
We have seen that a body surrounded by a fluid is
pressed upwards by a force equal to the weight of the
fluid displaced. This is the case with all bodies
weighed in air, and consequently their real weight
(i.e. their weight in vacuo) is greater than their ap-
parent weight by the weight of the air displaced
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48 Hydrostatics.
If two bodies, the volumes of which are not equal,
balance one another in air, their real weights are not
the same : that which has the greater volume is really
the heavier. Thus, let the volumes of the two bodies
be V and V\ their real weights W and W, Then,
if they balance in air, W^ Vs=: W— Vs, where s is
the specific gravity of air.
Hence, W^W' + { F- V% or,
^is greater than W, if Fis greater than V.
When the weights of two bodies, of small and
nearly equal volumes, are compared, the diflference
between their real and apparent weights is so slight
as to be in most cases of no practical importance. But
where the volume of one body is much greater than
that of the other, the quantity ( F— V')s cannot so
easily be neglected. Thus, in answer to the question.
Which is the heavier, a pound of feathers or a pound
of lead ? we should say that the apparent weights of
both are the same ; but the real weight of the feathers
is greater than that of the lead, and the two would not
equilibrate each other in vacuo.
§ 47. Bzamples.— (i.) A solid cube of metal the edge of
which is 3 inches, and whose specific gravity is 7, is wholly
immersed in water and is supported by a string attached to it.
Find its apparent weight in water.
Let Wh^ the weight of the body in air, which differs very
slightly from its real weight, A its apparent weight in 'wSiter,
and y the resultant vertical pressure acting upwards. Then the
body is in equilibrium under the action of these three forces,
and, therefore, W^ Y+A,
Now, the resultant vertical pressure K equals the weight of
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Fluid-pressure on Bodies immersed, 49
water displaced «3» xze;, where w=» weight of a cubic inch of
water, and W^ 3* x 7 x w.
Hence -4« W— y«3*x 7xw-3»xw = 6x3»xz(/
162 X 1000 ^ „ ^ -
° .1728 °'- = Slbs. i3|o^.
(2.) A body weighs in vacuo 560 grams, and in water
60 grams : find its volume.
Here W^ A = 560 - 60 = 500 grams a weight of water dis-
placed. Hence volume of the water displaced, ue., the volume
of the body = 500 cubic centimetres.
{3.) Two hollow spheres, the volumes of which are 100 and
200 cubic centimetres respectively, balance one another in
vacuo. What weight must be placed inside the larger that they
may balance in -water ?
Weighed in water the larger sphere will seem to be the
lighter, since the force supporting it is greater. Now the force
supporting the larger sphere in water is 200 grams, and the
force supporting the smaller sphere is ic» grams. Hence, for
equilibrium, the weight of the larger must be increased by 100
grams without increasing its volume. This may be done by
placing 100 grams inside.
{4.) Find the acceleration with which a heavy smooth body
will sink in a perfect fluid less dense than itself.
Let ff^-the weight of the body, 5 its absolute specific
gravity.
Let / = the specific gravity of the fluid.
Then the volume of the body is — - (§ 17) and the weight of
5
the fluid displaced is W—,
Hence, the resultant force measured in gravitatic*n units,
causing the body to descend, is W ( i— -^ and the mass of
the body moved is — .
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50 Hydrostatics,
Ifi therefore, / is the acceleration with which the body de-
scends, it is shown in books on Mechanics that
If the body be specifically lighter than the fluid, the body, if
placed under the fluid, will tend to rise to the surface, and
the acceleration /= ^-^ . g.
s
Exercises. V.
1. Find the weight in water of a piece of zinc (sp. gr. =7)
that weighs 8 oz. in vacuo. ^ *"
2. Find the volume of a body that weighs 350 grams in
vacuo and 225 in water. ' • u j ^ - T i J - liJ^t V^
3. A block of stone 2 cubic feet is wholly immersed in .
water. With what force is it buoyed up ? ". 3 • i '^ I 1 I Xf^tjb
4. A piece of metal weighs 36 lbs. in air and 32 lbs. in fresh
water. What will it weigh in sea-water the sp. gr. of which is
1-025? y-<jvj-. V -- ^-z
5. An air-ball, the volume of which is 6 cubic decs., weighs
in air 15 grams. Find its real weight, having given that the
weight of I cubic centimetre of air is 0*0013 grams.
6. A round disc of lead (sp. gr. = 1 1 '35) area 5 sq. centi-
metres and thickness 2 centimetres, is fixed to a piece of cork
(sp. gr. =o*24) of same area and 8 centimetres thick. Find the
weight of both in water.
7. The edge of a hollow cube of lead is 8 centimetres, its ,
tliickness is 2 centimetres. Find its weight in water. /'•^•*' ^^^ " '^''^
8 A bottle weighing 200 grams is completely filled with
water, when it is found to weigh 800 gr. If a piece of iron
(sp, gr. =7*2) is placed in the bottle, the bottle with its contents
weighs 955 grams. Find the weight of the iron.
9. A piece of metal of specific gravity 8, and weighing
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Fluid-pressure oft Bodies immersed. 5 1
20 lbs., is dropped into a cylinder Jilledmth water. Find the
additional pressure on. the base.
10. A cylinder of wood (sp. gr. «=o*75), one decimetre high,
with a sectional area of 16 sq. cm., is immersed with its axis
vertical in alcohol (sp. gr. =o*8). Find the least weight that
must be placed on the top of it to bring its upper surface on a
level with the alcohol.
11. A piece of wood weighing 8*25 grams (sp. gr. bo*66) is
placed 4 ft. 7 in. deep in water and is free to rise. N^lecting
all frictional resistance, find its velocity when it reaches the sur-
face of the water.
12. Find the time occupied by a stone (sp. gr. b3*2) in
falling from rest through 55 feet of water.
13. Two spheres whose radii are respectively I^ and r cm.,
and both of which are heavier than their respective bulks of
water, are of equal weight. What weight of metal (sp. gr. =j)
must be attached to the doUom of the larger that they may
balance each other in water?
14. Two masses of given specific gravities balance when
suspended from the equal arms of a lever in a known fluid.
What is the specific gravity of a fluid in which they balance
when one of the masses is doubled ?
15. A cubic inch of one of t\ro liquids weighs a grains, and
of the other 6 grains. A body immersed in the first weighs /
grains and in the second g grains. What is its reai weight and
what is its volume ?
VIII. F/oafing Bodies — Metacentre,
§ 48. When a body is immersed in a fluid, we
have to consider three separate cases : —
I. The weight of the body may be greater than
the weight of the fluid displaced, in which
case motion will take place in the direction of
the greater force, and the body if left to itself
will sink.
E 2 Digitized by Google
52 Hydrostatics,
2. The weight of the body may be equal to the
weight of the fluid displaced, in which case
it will rest anywhere in the fluid.
3. The weight of the body may be less than that
of the fluid displaced, in which case the re-
sultant vertical pressure will force the body
upwards, and it will float
These three cases can be easily illustrated by ex-
periments. The first of them has been already con-
sidered in the preceding paragraphs, and it has been
shown that in order that the body may be prevented
from sinking in the fluid, it must be upheld by a force
A equal to its apparent weight, such that
when ^is the weight of the body in vacuo, and Y
the resultant vertical pressure, or weight of the fluid
displaced.
The second occurs less frequently, and does not
now need to be separately treated. Later on, when
speaking of the formation of drops, we shall have
occasion to consider the form assumed by a mass of
fluid which is wholly supported by the external re-
sultant pressure.
The third case is of great practical importance,
and brings us to the consideration of floating bodies.
§ 49. Principle of Flotation.— If we take a body
the weight of which is less than the weight of an
equal volume of a liquid into which it is immersed,
the resultant vertical pressure will bring it to the sur-
face, and the body will be found to assume a posi-
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Principle of Flotation. 53
tion of equilibrium in which it will be only partly
immersed. If the experiment indicated in § 45 (i) be
tried with a body lighter than Fig. 24.
water, it will be found that the
weight of the liquid that escapes
in consequence of the partial im-
mersion of the body is exactly l^^^^^^^Sf
equal to the weight of the body ]|OI
itself. ' ^
The relation between the part of the body which
is immersed and the part that rises above the sur-
face of the fluid is determined by the principle that
the two forces acting on the body must equilibrate
each other ; ue. the weight of the whole body acting
downwards must equal the weight of the fluid dis-
placed acting upwards. Thus if a b (fig. 24) be the
intersection of the body with the surface of the liquid
in which it floats, the position of a b is determined by
the equation W =■ Y^ where ^ is the weight of the
body, and Fthat of the liquid displaced Moreover,
if G (fig. 24) be the centre of gravity of the body and
s the centre of buoyancy, g and s must be in the
same vertical line, for otherwise the body would be
acted upon by a couple which would produce oscil-
lation or rotation. Hence the conditipns of a body
floating in perfect equilibrium are : —
1. The weight of the body must equal the weight
of the fluid displaced.
2. The centres of gravity of the body and of the
fluid displaced naust He in the same vertical line.
§ 50. Stable and unstable Equilibrium. — If a
floating body be slightly displaced, so that the points
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54
Hydrostatics.
s and G do not lie in the same vertical line, the mo-
ment of the forces acting on the body may either
tend to restore the body to its original position, as in
fig. 25, in which case the equilibrium is said to be
stable^ or to overturn the body altogether, as in fig. 26,
in which case the equilibrium is unstable.
Whether a floating body, partly immersed, can
suflfer a small displacement without being overturned
is a matter of the greatest practical importance, as the
safety of all vessels at sea depends on it For, in
consequence of the action of the tides and waves, a
vessel seldom or never floats in perfect equiHbrium,
but is contmually undergoing a rotatory displacement,
which makes it oscillate about its original position of
equilibrium Now we see that if s is above G the
body will alv^ays return to its original position, or in
this case the equilibrium of the vessel may be said to
be stable. But where s is below G, the body is very
likely to overturn if displaced, and hence the danger
of overloading the deck, and the advantage of ballast
in lowering the centre of gravity of the vessel. But
a body may float in stable equilibrium even if s be
below G, as we shall see presently ; and consequently
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Stable and Unstable Equilibrium. 55
we cannot determine from the relative positions of s
and G only whether the equilibrium of a floating body
is unstable or not.
§ 51. Metacentre. — When a body floating in equi-
librium receives a slight displacement, the centre of
gravity of the fluid displaced assumes a new position
on \^hich the stability of equilibrium mainly depends.
Thus let ABC (fig. 27) be a body floating in equili-
brium, and let s be the centre of buoyancy. Suppose,
now, the body to undergo a slight displacement, in
Fig. 37.
consequence of which the centre of buoyancy changes
to s'. If, then, the vertical through s' meets in the
point M, the line through s and G which was vertical in
the first position of the body, it is clear that when m is
above g the equal forces acting on the body will form
a couple tending to restore the body to its former
position, and that when m is below G they will tend
to overturn the body. The point m is called the
metacentre of the body, and its position depends on
the shape of the body and on the position of its centre
of gravity. Of course, if g is below s the point m will
be above g. Hence it follows generally that the
equilibrium of a floating body is stable or unstable
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56 Hydrostatics,
according as the metacentre is above or below the centre
of gravity of the body.
The metacentre may be defined as the point in
which the vertical through the centre of buoyancy of
a floating body which has undergone a slight dis-
placement, intersects the line drawn vertically through
the centre of buoyancy in the original position of equi-
librium.
If a body floats wholly immersed in a fluid, no
change in the position of the body will alter the
relative positions of the centres of gravity and buoy-
ancy. Hence in this case the equilibrium cannot be
stable unless the centre of gravity is below the centre
of buoyancy. If the positions of these points be
reversed, the slightest oscillation will cause the body
to overturn.
§ 52. Bxamples.— (I.) A body the specific gravity of the
material of which is s floats in a liquid specific gravity s'. Find
what fraction of its volume will be immersed.
Nearly all problems on floating bodies may be solved by the
application of the principle of Archimedes, i.e, by equating the
weight of the floating body with the weight of the fluid it dis-
places.
Let Fequal volume of the body; V of the portion immersed.
Then K^ = weight of body; and F'/s: weight of fluid dis-
placed ;
V s
(2.) A cylinder the height of which is 12 inches floats two-
thirds immersed in water. Find what part of it will remain im-
mersed if a liquid the specific gravity of which is o*2 be poured
upon the surface of the water so as to completely cover the
cylinder.
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Fluid-pressure on Bodies immersed. 57
When the cylinder floats in the water only, the weight of
the cylinder equals the weight of the water displaced.
where s is the specific gravity and a the sectional area of the
cylinder.
,*, specific gravity of cylinder = §.
When the lighter liquid is poured on to the surface of the
water, the weight of the cylinder equals the sum of the weights
of the fluids displaced.
If z equals in this case the part of the cylinder immersed in
the lighter liquid we have
I2xf = 2X0*2+(I2 — 2)
or 80 = 22+ 120- IQ2
.', 2 a. 5 inches.
Hence the cylinder rises one inch out of the water in conse-
quence of the additional upward pressure due to the liquid
poured into the vessel.
(3.) Required the weight that must be attached to the end of
a straight rod of known weight and specific gravity, that it may
float vertically in water.
A uniform straight rod cannot float in a vertical position,
because the centre of gravity is always above the centre of
buoyancy. If, however, a heavy Fig. 28.
particle of no considerable magni-
tude be attached to the lower end
of the rod, the centre of gravity may
be lowered as much as we please.
What is required, therefore, is the
weight of a particle that will lower
the centre of gravity to the depth, at
least, of the centre of buoyancy.
Let G be the centre of gravity
of the rod A b, a its sectional area, and s its absolute specific
gravity, and lV\\s weight. Let x be the weight attached to the
end A, and g be the centre of gravity of ^and x. Then
w+a>
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58
Hydrostatics,
{IV+ x)kg=: JVx A G where ^= 2AGxaxs
Also, by principle of Archimedes, IV-k- x^AK,a
a
Now, in order that the rod may float vertically, Ag must no
be greater than A s. Put A^« A s. Then
H^ _ W+x
{IV+x)2as 2a
.'. ^^{iv+xy
This is the /easi weight that will suffice. Any greater weight
than this may be added to A, provided it is not great enough
to completely immerse the rod. Hence, the maximum weight
possible is x, where — = IV +x,
or x^fV
(7-)-
The weight required, therefore, must have a value some-
where between
^(^_,)a„d^(i-,)
In the above calculation the volume of the heavy particle, being
P small, has been altogether neglected.
(4. ) A uniform rod of given length
and specific gravity moves freely about a
point at a given height above the sur-
face of a liquid in which it is partly
immersed. Required its position of
equilibrium,
I>et A B be the rod ; PV, Y, G and s
as before.
The position of equilibrium is deter-
"^^-
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Fluid-pressure on Bodies immersed. 59
mined by equating the moments of W and Y about A. Tak-
ing the sectional area, which is uniform, as unity we have
ff^=ABxjand y=BKx^,
where s and d are the specific gravities of the material of the
rod and of the liquid respectively. Also
^xGM=yxSN /. if=^JL
Y GM
. -iLuL^Af by ^imilaiity of triangles.
BK. rf AG "^ ' ^
• AB S _ A B + AK _^AB
' ' AB-AK * d 2 * 2
•■•?-'-(^)'
•••rfvo-j) •••— 'vo-a-
If A c is given, A K can be expressed in terms of A c and
of the angle B a c, which is thus determined.
Exercises. VI.
1. A cylinder of wood (sp. gr. =0*6) and 12 inches high
floats with its axis vertical in water. To what depth will it be
immersed ?
2. A cube of oak (sp. gr. =097) each edge of which is
6 inches, floats partly in sea water (sp. gr. i '028) and partly in
olive oil (sp. gr. =0-915). Find what part of it is immersed in
each liquid.
3. A uniform block of metal 10 inches high (sp. gr. = 8) floats
in mercury (sp. gr 13*6). Plnd how much it rises out of the
mercury if water be poured on the surface of the mercury so as
to completely cover the block of metal.
4. A cylinder floats vertically in a liquid. Compare the
forces necessary to raise it and to depress it to an equal extent.
5. A heavy uniform rod weighing 3 kils. and sp. gr. = 6
moves freely under water in a vertical plane about a hinge at one
end. If a string tied to the other end supports the rod in a
horizontal position, find the tension in the string.
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6o Hydrostatics.
6. A piece of gold (sp. gr. 19*25) weighing 96*25 grams
immersed in a vessel full of water causes 6 grams of water to
be displaced. Is the gold solid ? If not, find the size of the
hollow.
7. Two equal globes, the volume of each being 100 c.c, are
suspended from the equal arms of a lever, the one hanging com-
pletely immersed in water, the other in a liquid of sp. g^. =0*8.
What additional weight is required to make them balance ?
8. A cube of metal is floating in mercury (sp. gr. = 13*6).
When a weight of 170 lbs is placed on the top, it is observed to
sink 3 inches. Find the size of the cube.
9. If an iceberg (sp. gr.= 0*918) float in sea-water, what
is the ratio of the part submerged to that which is seen above
water ?
10. Find what quantity of cork must be attached to a man
whose weight is 168 lbs. and sp. gr. = i -12 so as to enable him
just to float in water.
11. An iron ball of 12 lbs. weight floats in mercury covered
in water. Find the weights of the parts in the two fluids ; having
given specific gravity of mercury =13*6, specific gravity of iron
= 7-5.
12. The specific gravities of the upper and lower of two
fluids that do not mix are 0*9 and i*i ; the- upper fluid is 4 in.
deep ; a cube with an edge of i foot and sp. gr. 0*75 floats in
the liquids ; how much of it is immersed ?
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specific Gravity, 61
CHAPTER IV,
SPECIFIC GRAVITY, AND MODES OF DETERMINING IT.
IX. Application of the Principle of Archimedes to the
determination of the Specific Gravity of Bodies,
§ 53. To find the specific gravity of a substance we
require to know its weight in vacuo, and its volume,
W
since 7^ = ^f- The volume of the body, expressed m
cubic centimetres, is numerically equal to the weight
of an equal bulk of water at the standard tempera-
ture expressed in grams, since the weight of a unit-
volume of water is one gram ; and, therefore, the
specific gravity of a substance is expressed numerically
by the ratio of its weight to that of an equal bulk of
water.
Instead of the weight of the body in vacuo, we
generally substitute the weight in air, the difference
being unimportant for solid and liquid bodies of small
magnitude. Where the volume of water equal to that
of the body can be directly found, the specific gravity
is very easily determined ; but where this is not the
case different methods have to be employed.
The principle of Archimedfts tells us that a body
immersed in a fluid is pressed upwards bv a force
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62 Hydrostatic!:.
equal to the weight of the fluid displaced. Hence the
specific gravity of a body can be found by comparing
the real weight of a body with the difference between
its real weight and its apparent weight in water, since
this difference is equal to the weight of water dis-
placed. If, therefore, W represent its real weight,
and A its apparent weight in water, the specific
W
gravity of the body equals ^ .
§ 54. To find fhe specific gravity of a solid body
insoluble in water.
This can be determined roughly by an experiment
similar to that described in § 45 (i), where the weight
of the body may be directly compared with the
weight of the water displaced. But a more accurate
result may be obtained by using the hydrostatic
balance for finding A^ the weight of the body in
water. The specific gravity can then be determined
by dividing W^ the weight of the body, by W^A^ the
loss of weight in water ; or
specific gravity = -^ — -
% 55. To find the specific gravity of a solid body
that fioats in water.
In this case the body may be said to have a nega-
tive weight in water : Le, it requires the application of
a positive force to make it weigh zero when completely
immersed. Let ^ equal the weight of the body, P
the force required to completely immerse it, then the
force necessary to resist the resultant vertical pressure
= ^ + P = weight of water displaced by the whole
body.
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specific Gravity. 63
W
Hence, specific gravity =
W^-P'
What we want, therefore, is to determine P, This
we can do by attaching a heavy body called a sinker
to the light body, and by weighing them both in
water. If they weigh zero, the sinker's weight in
water is the force P required ; but if they tend to
sink and are found to weigh -5, then the sinker's
weight in water is too great by this weight B.
If, therefore, A equals the apparent weight of the
sinker in water,
A-^B^P,
and the specific gravity of the body is r^jz :; =.
fr + A — o
For determining these weights the hydrostatic balance
is again employed.
§ 56. To find the specific gravity of a liquid, by
weighing a solid in it.
Let the solid weigh W in air, and let a be its ap-
parent weight in the given liquid.
Then ^—dJ = weight of the volume of liquid
displaced by the solid.
Let solid weigh A in water.
Then W— A^ weight of the volume of water
displaced by the solid.
But these two volumes are the same ;
.•, specific gravity of liquid = "
§ 57. To find the specific gravity of a solid body
soluble in water, but insoluble in a liquid of known
specific gravity.
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64 Hydrostatics,
If the body is soluble in water, its weight in water
cannot be directly found, and consequently the
method of § 54 is inapplicable to this case. If,
however, the body is insoluble in some other liquid
the specific gravity of which is known, we have the
means of determining the specific gravity of the
body.
Suppose the weight of the body to be W^ and its
weight in the known liquid a^ then W — a \% the
weight of the liquid it displaces ; and if s be the spe-
cific gravity of this liquid, — ^^ = the volume of
• s
liquid occupied by the body. It follows, therefore,
W—a
that the weight of an equal volume of water is ,
since specific gravity of water is unity, ^
W—a TVs
Hence, specific gravit}' of body= JV^ —
W-a
§ 58. Specific Oravity of Oases. — In determining
the relative weight of gases, air at 0° C, and at the
ordinary atmospheric pressure is taken as the standard
substance. The process is attended with so many
difficulties, owing to the peculiar properties of gases
which have not yet been considered in this work, that
many precautions are needed in order to obtain accurate
results. In comparing the weights of equal volumes
of any gas and air, it is necessary that the gases should
be subjected to the same atmospheric pressure, and
should be brought to the same temperatiure.
If we suppose JV to be the weight of a glass globe
full of air, and w its weight when empty, and if fV'
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specific Gravity, 65
be the weight of the same globe filled with any other
gas, then the specific gravity of this gas as compared
with air is
W— w
If the globe be filled with water from which all
air-bubbles have been carefully excluded, and if Wx
be its weight so filled, the specific gravity of the gas
as compared with water would be -— .
^ Wx—w
In determining with accuracy the specific gravity
of any substance, the temperature at which the experi-
ment is conducted must be considered; for it is
known that bodies generally expand when heated,
and the weight of the same volume of water, or of any
other substance, consequently varies at different tem-
peratiures. This subject is more fully considered
in treatises on heat
Before working the following exercises the student
is recommended to take pieces of brass, tin, zinc,
marble, wood, and other easily obtained substances,
and to determine their specific gravities by the hydro-
static balance, comparing the results with those given
in the tables.
Exercises. VII.
1. A solid soluble in water but not in alcohol weighs 346
grams in air and 210 in alcohol. Find the specific gravity of
the solid, that of alcohol being 0-85.
2. A solid weighs 100 grams in vacuo, 85 grams in water,
and 88 grams in another fluid. What is the specific gravity of
the fluid?
F
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66 Hydrostatics.
3. A piece of copper weighs 31 grains in air and 27*5 in
water. Find its specific gravity.
4. A piece of wood (sp. gr. ==074) of 32 cubic inclies floats
in water. How much water will it displace ?
5. A body floats in water with one-eighth of its volume above
the surface : determine its specific gravity. How much of it
will be submerged in a fluid whose specific gravity is 09 ?
6. A body floats in a liquid (sp. gr. =si3*5), ^^^%. ^^ ^^s
volume is above the surface : find specific gravity of the body.
7. If a ball of platinum weigh 21*4 oz. in air, 20*4 oz. in
water, and 19-6 in hydric-sulphate, find the specific gravity of
the platinum and of the acid.
8. A piece of marble, specific gravity = 2*84, weighs 92 grams
in water and 98*5 grams in oil of turpentine. Find the specific
gravity of the oil.
9. A piece of cork weighs 2 oz., and its specific gravity is
0-24. What is the least force that will just immerse it?
10. A piece of iron, sp. gr. 7*21, and weighing 360*5 grams
is tied to a piece of wood weighing 300 grams, and the weight
of both in water is 110*5 grams. Find the specific gravity
of the wood.
11. A cubic block of wood each edge of which is 8 cm. is
put into water. If the specific gravity of the wood is 0*85, with
how much more wood must it be loaded, so that its upper sur-
face may sink to the level of the water ?
12. A cylinder of wood 20 inches in height, and a cylinder of
lead I inch in height, are united so as to form one cylinder
21 inches in height, which is found to float in water with
3 inches projecting above the surface. Find the specific gravity
of the wood, that of the lead being 1 1 -4.
X. Ot^er Methods of obtaining the Specific Gravity of
Substances. — Hydrometers.
§ 59. The Specific Gravity Bottle.— This is a
bottle made to hold a certain weight of water at the
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The Specific Gravity Bottle. 6^
standard temperature. It may be employed for deter-
mining the specific gravity of a liquid or a powder.
1. Specific Gravity of a Liquid, — Suppose the bottle
when empty to weigh w grams, and that it holds
I GO grams of water. Let the bottle be filled with
the liquid the specific gravity of which is required,
and suppose it then to weigh (91+ a/) grams. Then
the weight of the liquid in the bottle is 91 grams,
and the weight of an equal bulk of water is 100
grams.
Hence, specific gravity of liquid is -^ = 0*9 1.
2. Specific Gravity of a Powder, — Let the powder
be first placed in the bottle, and let the bottle be then
filled with water, and suppose the weight of the
contents of the bottle, /.^., of the powder and water,
to be K, Then if ^be the real weight of the powder,
and y the weight (unknown) of the water it displaces,
^^=100+ w- y,
and .-. Y^W-^-ioo-K
W
/. Specific gravity of powder = -j^
Take 4 grams of powdered glass, put it into
the bottle, and fill with water ; then the bottle with
its contents will be found to weigh 102*5 + 0/ grams ;
/• F=4— 2-s=i-5; .•, specific gravity=Ttj=2-6.
§ 60. Hydrometers.— The hydrometer^ consists
essentially of a straight stem, loaded at one end, so as
* This instrament is said to have been invented by Hypatia,
the daughter of Theon Alexandrinus, who flourished about the
end of the fourth century ; though there is some foundation for
the opinion that tht invention is due to Archimedes.
F 2
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68 Hydrostatics,
to float in a vertical position. By observing the
depths to which it sinks in two different Hquids, the
relative weights of these two liquids can be deter-
mined. Hydrometers are of very different forms, and
may be used for finding the specific gravity of liquids
or of solid bodies. We shall describe two varieties
only.
§ 6i. Common Hydrometer. — ^This instrument,
invented by Fahrenheit,* consists of a straight stem,
Fig. 3a usually made of glass, which terminates in two
^ hollow spheres. The lower sphere is loaded
^ with mercury, so that the instrument may
float in a vertical position.
Let a = section of the stem.
w = weight of the instrument.
V = volume „ „
Suppose the instrument to float in water with *
the point d of its stem on the surface of the
water, and in some other liquid, the specific
gravity of which is to be found, with the point c on the
surface.
Then, if j be the specific gravity of the liquid, it
follows from the principle of Archimedes that
'w = si^—a X A c) = v—a x a d
taking the weight of the unit-volume of water as
unity.
• __ Z^~gXAD
§ 62. Nicholson's Hydrometer.' — This form of
instrument enables us to determine the specific gravity
> 1724. * 1787.
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Hydrometers, 69
of solid as well as of liquid bodies. It consists of a
hollow body, b, connected by a fine stem with a small
dish, A, at its upper end, and at its lower end with a
cup, c, loaded so as to ensure equilibrium.
When floating in distilled water, with a certain
weight, k^ in the upper dish, the instrument sinks
to such a depth that a fixed mark, 0^ is on the surface
of the water.
1. If we want to find the specific gravity of a given
liquid, we place the instrument in it,
and place weights in the upper dish ^^^' 3''
till the fixed mark, ^, is on a level with
the surface of the liquid.
If, then, W be the weight of the
instrument, k the weight added to sink
it to in water, and w be the weight
added to sink it to in the given liquid,
W-^-k is the weight of the water dis-
placed ; and W-^-w is the weight of
the liquid displaced ; and, as the same
bulk of fluid is displaced in each case,
the specific gravity of the liquid is
W+k'
2. To find the specific gravity of a solid.
Let k be the weight as before that must be
placed on the instrument to bring the point ^ on a
level with the surface of the water. Place a small
piece of the solid on a, and diminish the weight k^ so
as to keep the instrument at the same level.
Let m be the weight now employed.
Then k—m=si weight of solid body.
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yo Hydrostatics.
Now place the substance the specific gravity of
which is to be found in the cup c, in which case a
weight greater than m by the weight of the water
displaced by the body must be placed on a to main-
tain the instrument in the same position.
Let n equal the weight now in a.
Then «—f« = the resultant vertical pressure on the
solid in c,
= the weight of the water displaced by it
Hdnce, specific gravity is -— ^.
Exercises. VIII.
1. A piece of glass the weight of which is 50 grains is placed
on the upper dish of a Nicholson's hydrometer, and it is found
that an additional weight of 235 grains is required to sink the
instrument to the fixed level. If, however, the glass be placed
in the lower cup a weight of 250 grains is required. Determine
the specific gravity of the glass.
2. A globe of glass holds 5 litres of water at standard tem-
perature. When full of water it weighs 5500 grams. "When
full of air it weighs 506 "465 grams, and when full of hydrogen
it weighs 500*447 grams. Find the specific gravity of air
as compared with water, and of hydrogen as compared with air.
3. A bottle filled with water is found to weigh 500 grams.
If 180 grams of powder are introduced into the bottle, the
bottle with its contents weighs 575 grams. Required the
specific gravity of the powder.
4. Into a specific gravity bottle capable of holding looo
grains of water, 300 grains of a certain powder are introduced.
The bottle is then filled up with a liquid, specific gravity o*8,
and the contents of the bottle are found to weigh 980 grains.
Find the specific gravity of the powddr.
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The Motion of Liquids. 7 1
CHAPTER V.
THE MOTION OF LIQUIDS.
XI. Liquids moving by their own weight,
§ 63. It is proposed to consider in this chapter a
few of the most elementary propositions connected
with the movements of fluid bodies. Very little can
be attempted without the application of some of the
higher processes of mathematics, and we shall endea-
vour, therefore, only to indicate the principles of the
methods on which the solution of this class of pro-
blems depends. As all liquids are more or less viscous,
absolute agreement cannot be expected to exist
between the conclusions arrived at on the assumption
of a frictionless fluid and the results of direct ex-
periments.
§ 64. DefinitioD. — The vertical distance between
the surface level of a liquid in a vessel and the centre of
the orifice through which it escapes is called the head
ox pressure height under which the flow takes place.
§ 65. Torricelli's Theorem. — The velocity with
which a liquid escapes from a small hole in the bottom
or side of a vessel was the subject of numerous experi-
ments made by TorricellL^ The result at which he
* Born in Italy in 1608 ; died in 1647.
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72 Hydrostatics.
arrived was that the rate of efflux was the same as
would be acquired by a body falling freely from the
surface level of the liquid to the centre of the orifice
from which it escapes.
Let A B c D be a vessel having a small open-
ing, o, in one of its sides. Let the area of this open-
Fig. 32. ing, which is supposed to be very
small, be ^, and suppose a small mass
of liquid, ;;/, the thickness of which is
Cy to occupy this opening.
Then if / be the pressure in excess
of the atmospheric pressure urging
this mass forwards, measured in units
of force per unit-area, p=^g. da h
where d is the density of the liquid, and h the pressure
height, o H ; and the work done by this force p in
urging the mass m from a position of rest through
the distance c is
p X c =i g, dah.c)
and since the work done is equal to the energy which
the mass m acquires,
p , c '=' , where v is velocity of efflux ;
sr.d ahc=i dac —
^ 2
or v^ -=■ 2 g h
/>., the velocity is that which would be acquired by
the mass m in falling freely from the surface level of
the liquid to the centre of the orifice.
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TorricellVs Theorem, 73
In this result each particle of the mass is supposed
to move at right angles to the area of the orifice, and
to escape into the air at a pressure equal to that at the
surface level
§ 66. Experimental Test.— The accuracy of the
results given by Torricelli's theorem may be tested
for different liquids, by fig. 33.
taking a vessel having
apertures in one of its
sides through any of
which the water or other
liquid it contains may
escape. If the vessel
be kept constantly full,
and one of these aper-
tures be opened, the « 6 c
liquid will flow through
it, and will descend in a curve similar to the path of
a body projected horizontally from a certain height
above the ground.
If the distances payjf>b,pche observed, the accu-
racy of Torricelli's theorem can be tested.
For, let the height o c = ^, then according to the
theorem, if v be the velocity of the liquid at c,
V = -^ 2gh, and the velocity will be uniform and
horizontal.
If then / equal the time occupied by a particle in
moving from c to c, we know by the second law of
motion that / equals time occupied by a body falling
freely from c to /
= time of describing/^ with the uniform velocity v.
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74 Hydrostatics.
Hence, pc^='tv\ also cp = — ^
orvj=pcx^/ S ^
V 2C/
Now, if the distance/ c be known, and if /^ be ob-
served, the value of z/ so found can be compared with
that given by the formula z/ = n/ (2^. o c). On
comparing the results, obtained by experiment, with
those obtained from Torricelli's theorem, a certain
difference will be always found to exist. Some of
the reasons of this difference, apart from those depen-
dent on the viscosity of the liquid, will be considered
later on.
§ 67. Belation of Telocity of Flow to Sectioiial
Area of Vessel* — Let a b c d be a vessel, which is
Fig. 34.
kept constantly full of liquid. We may consider the
mass of the fluid, which is supposed to be incom-
pressible, to be divided into small laminse moving
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Vena Contracta. 75
parallel to one another and of uniform sectional area
for very small differences of depth. Thus, libcde,
p^rsyhe two small laminae, we may suppose bc^=^de
and/^ ^^ r s. Let each particle in the section be
have the velocity, Vy and each particle in the section
pq the velocity v'. If, then, the section be descend
to d e in one second, the section / q will descend
to some depth, rs, in the same time ; and the volume
bcde will be equal to the volume /^rj, since the
quantity of liquid between de and /^ remains con-
stant
If, then, A equal the section be ox de, and a' the
section/^ or r J,
A z/ = a' z/' = volume discharged in one second,
/. V \ v' \\ a! : A,
or the velocities are inversely proportional to the
sectional areas.
§ 68. Vena Contracta.— When a liquid issues
into the air from a small opening in a thin plate, the
stream is found, first of all, to converge, so that it
contracts rapidly for some little distance from the
orifice. This fact may be very easily observed, and will
be seen in all cases such as those shown in figs. 33, 34.
The area of the jet at its narrowest part is known
as the vena contracta^ and was first made the subject of
investigation by Sir Isaac Newton. It is found to
be generally about three-fifths of the aperture of the
orifice, but varies with the shape of the aperture and
the pressure height. The value of this fraction, which
is called the eoefficient of eonir action ^ can be determined
by experiments only. In all calculations with respect
to the rate of the emptying of vessels, the method
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jS Hydrostatics.
usually adopted has been to apply the results of Tor-
ricelli's theorem, and to substitute the area of the vena
contracta for that of the orifice. This method, we
shall see, gives us a merely empirical result, and is
faulty in so far as it overlooks some important elements
in the problem.
§ 69. MeasTire of Pressure on the Walls of a
Pipe. — Suppose a liquid is flowing through a pipe
A B, the pressure which it exerts on the walls of the pipe
at any point can be experimentally determined by in-
serting gauge glasses, />., thin glass tubes of about
Fig. 35.
f of an inch diameter, into the pipe at those points
at which the pressure is to be found. Since fluids
transmit their pressure equally in all directions, the
forward pressure exerted by the moving liquid will be
equally exerted on the walls of the pipe, and will
force the liquid up the gauge glasses to a height corre-
sponding to the pressure produced. If the pipe is of
uniform sectional area the pressure produced at all
points will be the same, friction being neglected, and
the liquid in the pipe will rise in the gauge glasses
to the same height at all points in the pipe.
§ 70. Eolation of Pressure to Sectional Area of
Pipe. — If the sectional area of the pipe varies, the
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The Motion of Liquids through Pipes. Jj
pressure is no longer uniform, but is found to be
greatest where the area is greatest, and vice versd.
Suppose the liquid to be moving with a uniform
velocity in that part of the pipe a b (fig. 35) which
has a uniform sectional area. If, then, we consider an
element a of the liquid, we see that the pressure on
either side of a must be the same, since any increase
of pressure from behind would cause it to move with
an acceleration, which is not supposed to be the
case.
Suppose now the element to have reached b, and
to be entering the wider part of the pipe, its velocity
in the direction of the pipe's length will now be less
than it was before, and will continue to decrease as
the pipe widens. Hence the pressure of the liquid
behind the element a urging it on must ever be less
than that in front of it resisting its advance, and con-
sequently the pressure must increase with the area
of the pipe and with the decrease in the velocity of
motion.
After passing c, the widest part of the pipe, the
velocity begins again to increase, and as the resist-
ance to the forward motion of the particles must
consequently have become less, the pressure is also
diminished.
These results can be practically illustrated by
observing the height of the liquid in the several gauge
glasses placed at different points in the pipe. Due
allowance must be made for the frictional resistance
which causes the liquid to rise to a less height in the
glasses than it otherwise would, and gives a difference
between the theoretical and actual results which in-
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7S
Hydrostatics.
creases with the length of pipe traversed. This re-
sistance being neglected, we see that the pressure
which a liquid exerts on the sides of a pipe through
which it is passing varies directly with the sectional
area, the velocity at the two ends of the pipe being
the same.
If/i and/z be the pressure exerted at points where
the area of the pipe is a^ and azy we have
A :/2::«i :«2-
§ 71. Belation of Velocity of Flow to the Pres-
sure produced. — We have seen that in the case of a
steady flow through a pipe of varying area, the velocity
at different points varies inversely with the area, fric-
tion being neglected. We have now to consider how
the pressure changes with the velocity. This may be
experimentally exhibited by fixing the larger end of a
tapering pipe into a vessel of water kept constantly
full, the upper surface having only the ordinary at-
mospheric pressure upon it. If the pipe be furnished
FiQ. 36.
with gauge glasses, the pressure at different points
can be observed. The hydrostatic pressure at a is
evidently zero, or the same as that of the air into
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The Motion of Liquids. 79
which it issues, but the liquid will be found to stand
at a continually higher level in the gauge glasses as
the pipe widens towards the vessel. The difference
between the surface level of the water in any one of
the gauge glasses and that of the vessel is called the
fall of free level dX that point, and this fall is equal to the
difference between the height of the column of liquid
representing the hydrostatic pressure at a point in the
stream, and the pressure-height at the same point,
supposing the liquid to be in equilibrium.
Consider now a small mass m of the liquid in the
section of the stream at any point b, and let the
section of this mass be equal to a unit of area, and
its length equal to ^, so that it contains c units of
volume. Then if/ be the actual pressure in units of
force per unit of area at b, and if P be the pressure
at the same point due to the depth of the mass m
below the free surface l l' d of the liquid, then the
whole amount of work which the mass m has received
from the pressure behind it in moving through its
own length from a position of rest is -Px r, and since
/ is the pressure it is exerting, the work which it gives
to the liquid in front of it is/ x c. Hence the excess
of work which it receives is {P—p) c.
Now if JjTis the vertical depth of m below l l', and
h the height of the liquid in the gauge glass above niy
P=:gHdzxidp — gdh,
where d is the density of the liquid,
.-. {P^p)c^gd{H^h)c.
But the amount of work gained by the mask m \^
equal to its energy.
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8o Hydrostatics.
Hence, g d^H—h) c = ^^, where v is the velocity
of mass m.
And since m = dCy we have
Z;2 = 2g {H—h) ^ 2gDG
or, the velocity at any point in the stream is that due
to the difference between the statical and actual pres>
sure-height of the moving particles. Hence, m steady
flow the velocity generated from rest is that due to the
fall of free level.
If we suppose the sectional area at b to be twice
that at A, and at c twice that at b, and so on, then if
V be the velocity at a, the velocity at b = ^, since the
velocity varies inversely with the area in cases of
steady flow \
and since z/^ = 2 ^ . a d,
we have — =• 2^.dg, orBG = ^AD;
4 4 ' '
and — = 2 ^. D H, or c h=— ^ a d, and so
16 16
on, which gives the relation between the actual hydro-
static pressure at a point, and the velocity with which
the stream is passing that point
§ 72. Application to the Flow of a Liquid through
an Orifice in the Base or Side of a Vessel.— When a
liquid flows through an orifice, as previously con-
sidered in the case of Torricelli's theorem, the flow
does not actually take place in a direction at right
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Flcnv of Water through Small Orifice, 8i
angles to the orifice, at all points of the orifice, but
va directions such as are shown in fig. 37. The
nature of the flow is therefore
^'^' 37' somewhat similar to that indi-
cated in the preceding section,
in which the varying area of
the pipe is represented by the
space included between the
several stream lines. At all
points along the margin of
the orifice the direction of the
motion is tangential to the
plane of the orifice, and the water comes at once into
contact with the atmosphere. But at all intermediate
points the issuing stream exerts a pressure greater than
that due to the atmosphere, and consequently the
velocity at any of these points is not that due to the
height of the free surface of the liquid in the vessel,
but is that due to the difference between this height
and that of the column of liquid corresponding to
the actual pressure at the particular point in the
moving stream. This difference of surface level
has already (§ 71) been referred to as the fall of
free level ; hence the velocity of any particle in
the issuing stream is that due to the fall of the
free level \ or, if h equals the depth of a particle in
the stream below the free surface of the liquid, and
z equals the height of the column of liquid represent-
ing the pressure at that particular point, then if v be
the velocity with which this particle is moving,
(I = s/zglh—z).
In calculating the actual discharge in a given time
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82 Hydrostatics.
through a small orifice in the base or side of a vessel,
the following principles, which are essential conditions
of the flow, have to be considered :
1. The absolute velocity at any point in the plane
of the orifice is not that due to the vertical distance
between the point and the free surface level of the
liquid, since everywhere throughout the area, except
along the boundary of the orifice, the liquid is
under a pressure greater than that of the atmos-
phere, and consequently, as we have shown above,
the velocity is correspondingly diminished, or
V s= \/ 2 g(/i — z),
2. If the orifice be divided into small horizontal
bands, we cannot suppose the velocity to be constant
throughout each band, since the direction of the
motion of the fluid is different at different parts.
3. If we consider any vertical section of the liquid
as shown in the figure, since the direction of the mo-
tion towards the orifice is nowhere perpendicular td
the orifice, except at its centre, the actual velocity of
efflux is only the component of the whole velocity
as determined by the first of these principles, acting
in a direction at right angles to the plane of the orifice.
These considerations show how very complicated
is the problem of determining the actual discharge o
a liquid, per unit of time, through a given orifice, and
serve to indicate some of the causes of the discrepan-
cies between the results of actual experiments and
those given by Torricelli's theorem. Ordinarily, when
the orifice is small, the discharge is roughly calculated
by considering the velocity of efflux to be that due to
the distance between the centre of the orifice and the
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Flow of Water through Small Orifice. 83
level of the free surface, and by then substituting the
area of the vena contracta for that of the orifice. Thus
if a be area of a very small orifice, and y the co-
efficient of contraction, and if h be the vertical dis-
tance of the centre of the orifice from the free sur-
face level, the discharge per unit of time is taken as
y a ^/2 g h. But this result is not only a mere
empirical result, the degree of accuracy of which de-
pends on the value of y as experimentally determined,
but it leaves out of consideration some important cir-
cumstances under which the flow actually takes place. ^
§ 73. Effect of Friction on the Pressure of a
Liquid Flowing Vniformly througli a Pipe.— Hitherto
we have taken no account of the frictional resistance
which retards the passage of a liquid through a pipe,
and which, apart from all other circumstances, causes
the velocity of efflux to be less than that due to the
entire pressure-height above the orifice from which
the liquid escapes into the air.
The effect of this frictional resistance on the pres-
sure exerted by a liquid in flowing through a pipe of
uniform sectional area may be conveniently illustrated
by the following arrangement.
H A (fig. 38) is the section of a cylindrical zinc vessel
about 3 ft. high, into the side of which, near the bottom,
is fixed a brass pipe of uniform bore. At equal dis-
tances along this pipe are openings into which glass
gauge-glasses are fitted.
If the opening e be closed, and the apparatus
filled with water, the liquid will stand at the same
> *0n the Flow of Water through Orifices.* By Prof.
James Thomson. Report of British Association, Glasgow, 1 8 761
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84 Hydrostatics.
level in the vessel and in the pressure-tubes ; but as
soon as the end e of the tube is opened, the water
falls to different levels in the several tubes, being
lowest in the tube nearest the end, and rising by equal
differences in the several tubes. If the water in the
vessel be kept at the same level, it will be observed
Fig. 38.
that the surface levels of the liquid in the several tubes
lie on a straight line drawn through e to some point e'.
This line is found to be more nearly horizontal, the
greater the sectional area of the pipe through which
the water flows.
Now it is evident that if the liquid diiring its flow
had encountered no frictional resistance, its velocity
on leaving the vessel a h would have been that due to
the pressure-height above it, and its pressure on the
tube would not have exceeded that of the atmosphere
into which it issued. The additional pressure on the
tube is due, therefore, to the frictional resistance
encountered by the liquid ; and as the resistance to
be overcome at any point of the tube increases with
the distance of that point from the orifice, the pressure
of the liquid diminishes as it approaches the open end.
Let a equal the sectional area of the pipe, and let c be
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Effect of Friction. 8 5
the length of an element of liquid occupying it Let/
be the measure of the frictional resistance per unit of
length of pipe, and let p and/' be the actual pressures
per unit area at either end of the element c. Then if
we suppose m to be the mass of this element, and %f
its velocity, which is constant, — represents the
2
energy of this mass at all points in the pipe.
Now if the mass m moves through a distance r, the
work which it has received \^ pa x ^, and the work
which it has given to the liquid in front of it \^p' a x c.
Hence the resultant work gained in moving through
the space c\^(p a — p' a) c. But since the energy of the
mass remains constant, this amount of work is em-
ployed in overcoming the frictional resistance encoun-
tered by the liquid in its passage through the pipe.
The work done against friction is/ x c\ and
.% (p—p')ac^fy.x\
and since / — Z' ^=^gd(h—h')
where h^ h' are the heights of water required to pro-
duce the pressures / and ^ respectively, and d is
density of the liquid, we have
gda
which shows that the difference of pressures (as mea-
sured by height of the liquid in pressure-tubes) for
equal distances along a pipe of uniform sectional area
is constant.
It is to be here observed that whilst the kinetic
energy of the flowing water is constant, there is a
continual decrease of potential energy much in the
same way as when a body slides with a uniform
velocity down a rough inclined plane* zed by Google
86 Hydrostatics,
XII. Capillarity.
§ 74. In the preceding sections we have treated of
the equilibrium and motion of liquids in mass, but we
shall now consider a number of phenomena which are
mainly due to the action of forces on the molecules
of a liquid, and which are not immediately explicable
by the principles already stated These will be dis-
cussed under the two heads of Capillarity and Diffusion,
and they comprise the formation of drops, the relation
of liquids and solids in contact, and the intermixture
of liquids, either in contact with one another, or
separated by porous membranes. In discussing these
subjects we shall in each case commence with a few
experiments, and then indicate the nature of the prin-
ciples which seem to explain them.
§ 75. Drop Formation. — ^Experiments.— Take a
vessel containing olive oil and carefully drop into it
some small portions of water. These will be seen to
assume a spherical shape as they descend slowly
through the oil Raindrops are spherical; but the
rapidity with which they fall through the air is so
great that we are unable to see them long enough to
distinguish their form.
To show that a large amount of liquid will assume
a spherical shape if left free to the action of its inter-
nal forces, mix together alcohol and water, in such
proportions that the mixture may have tho density of
olive oil This will require about three-parts of alco-
hol to one of water. Then pour gently, so that it
may not be broken into parts, some olive oil into the
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Formation of Drops, 87
mixture, and it will be found to float in the form of a
sphere in any part of the surrounding liquid. With^
out any great difficulty globules of oil may in this way
be obtained, having a diameter of four or five inches.
These can be flattened between two glass plates, or
receive an indentation by being touched with a glass
rod, without losing their cohesion ; and as soon as the
external body is removed, they recover like an elastic
ball, their spherical form.
If pure quicksilver be scattered on a level surface
of glass, the smallest of the drops will differ very little
in form from perfect spheres. The larger drops, owing
to their weight, will be considerably flattened ; and if
a greater quantity cohere together, the tendency to a
spherical shape is visible, only in the somewhat curved
form of the upper surface.
In blowing an ordinary soap-bubble we obtain a
very good idea of the nature of the forces that act on a
fluid surface. For this purpose * take some common
soapsuds, or a Plateau's mixture of soap and glycerine,
and blow a small bubble at the end of a tube with a
bell mouth. A tobacco-pipe will answer if the bore
of the tube is large enough. After blowing the bubble
at one end of the tube, place the other end near the
flame of a candle. The bubble will contract and
drive a current of air through the tube, as may be seen
by its effect on the flame.
*This shows that the bubble presses on the air
within it, and is like an elastic bag. To enlarge the
bubble by blowing into the tube, work must be done
to force the air in, because the pressure inside the
bubble is greater than that of the air outside. This
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88 Hydrostatics,
work is stored up in the film of the soapsud, for it is
able to blow the air out again with equal force.'
§ 76. Surface-Tenrion.— -These experiments show
that when two fluids are in contact with each other
and do not mix, the surface separating them is in a
state of tension similar to that of a membrane
stretched equally in all directions, and that the
surface-particles have a tendency to approach one
another like those of the outer side of a bent watch-
spring. This contractile force, or surface-tension,
resides in the thin film separating the two fluids, and
does not extend to any appreciable depth below the
surface ; and to its action is due the spherical form
of the drop and of the soap-bubble. In the case of
the soap-bubble the superficial tension may be mea-
sured by considering the work done in order to pro-
duce a film of a certain area \ and the measure of the
tension per unit length is found to be the same as the
numerical value of the work done divided by the area.*
This tension seems to be due to the fact that
whilst the particles in the interior of a fluid are sub-
jected to equal forces on all sides, those on the sur-
face separating the two fluids are under the influence
of different molecular attractions, the result of which
is to give to the bounding surface a tendency to con-
tract. The surface-tension depends on the nature
and temperature of both media, decreasing as the
temperature rises, and vanishing altogether at that
temperature at which there is no longer a well-
marked distinction between the liquid and gaseous
states.^ Between any two liquids that do not mix,
> C. Maxwell, Theory of Heat, pp. 281, 283. - § 4.
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Surface-tension, 89
between a liquid and its own or another vapour,
and between a solid and a fluid of any kind, there
is a definite surface-tension, the value of whigh
for a unit of length is called the co-efficient of superfi-
cial tension, or co-efficient of capillarity, as having been
first considered in connection with the ascent of
liquids in capillary tubes. This tension may be
reckoned in c. g. s. units of force per linear centimetre,
and it is found that at a temperature of 2o°C. the
tension of a surface of water in contact with air is
81, that of mercury and air 540, that of alcohol 25*5,
and of olive oil 36'9 units of force. In fact, of ordi-
nary liquids water is found to have the greatest sur-
face-tension.
§ 77. Capillary Elevation and Depression. —
Experiments. — (i.) If a sheet of clean glass be im-
mersed in water and then removed, the water will be
found to have wet the surface of the glass. If, whilst
partially immersed, the surface of the water in contact
with the glass be observed, it will be found that the
surface of the water in the immediate neighbourhood
of the sheet of glass is raised. If we perform the
same experiment with mercury instead of water, we
„ shall find that the mercury
Fig. 39. Fig. 4a , , , -^
does not wet the glass,
and that the surface of
the mercury in the neigh-
bourhood of the glass will
be depressed.
If A B be the surface
of the liquid, c d the vertical section of the plate of
glass, the water (fig. 39) will be found to rise to some
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90 Hydrostatics,
height a b^ and the mercury (fig. 40) to be depressed
to the level a' b\ Moreover, the surfece of the water
in contact with the glass will be concave, and that of
the mercury convex \ and the elevation of the water
will be found to be somewhat greater than the
depression of the mercury.
(2.) Take now two plates of glass and let them ap-
proach each other in water. When they are suffi-
ciently close, the water will begin to rise between them,
Fig. 41.
and the height to which it rises will increase as the
distance between them lessens. Moreover, the sur-
face of water between the plates will be concave. If
the same experiment be made with mercury instead of
water, the mercury will be depressed between the
plates, but to a less extent, and the surface will be
convex.
If this experiment be varied by taking plates of
different thickness, the same results will be obtained,
thereby showing that whilst the amount of ascent or
depression depends on the distance between the plates,
it is independent of their thickness.
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Capillary Phenomena.
91
Fig. 42.
(3.) If instead of plain surfeces parallel to one
another we take two plates
of glass inclined at an angle,
and immerse them partly in
water, the water will rise be-
tween the plates to different
heights, and the intersection
of the surface of the water
with the plate will form a
curve, which is known as the
rectangular hyperbola.
(4.) If glass tubes of small bore, called capillary
tubes (from capiila, a hair), be partially immersed in
water, the liquid will rise in the tubes, and the height to
which it reaches will be found to vary inversely with
the diameter of the tube. This result holds good
whether the experiment be made in air or in a vacuum,
but the elevation is foimd to diminish as the tempera-
ture of the water increases. If mercury be used
instead of water, the liquid will be depressed, as pre-
viously stated in the case of parallel plates. The
curved surface of the liquid is called a meniscus^ and
the meniscus is concave for water and convex for
mercury.
§ 78. Principles of Capillarity.— When a liquid
comes in contact with a solid, the particles of the
liquid on the common surface of the two substances
are to some extent in a similar condition to those
of the free surface of a liquid ; />., they are not
equally attracted on all sides, as is the case with
particles in the interior. The interaction of the
molecular forces of the surface of the liquid and solid
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92 Hydrostatics,
gives rise to a tension in the common surface that
separates them. When a solid is in contact with two
fluids there is a different tension in each of the
surfaces separating a pair of media. In this case,
in order that the three tensions acting at any point
in the Hne of intersection of the surfaces may be in
equilibrium, their directions must be such that they
can be represented by the sides of a triangle, and
any one must be greater than the difference between
the other two. Now we have seen that the surface
of a liquid in contact with a solid is curved, and
for any given liquid in contact with a given solid
there is a definite angle of contact, called the angle of
capillarity^ which is cunite in the case of mercury, and
obtuse in the case of water in contact with glass. This
inclination of the surface of a liquid to that of a solid
at the margin of contact is due to the excess of the
tension of the surface separating the two fluids (for
example, air and water, or air and mercury) above
the difference of the tensions of the surfaces separating
the solid from each of the fluids.
§ 79. Height of ElevatioiL or DepressioiL of
Liquid in Capillary Tubes. — Let T denote the mag-
nitude of the superficial tension per unit length of
the free surface of the liquid inclined at a given angle
to the solid in contact with it. Then the vertical
component of T acting upwards or downwards is
the force tending to raise or depress the liquid.
Let / be the vertical component of T', and let r be
the radius of the tube, then 2 xr/ denotes the magni-
tude of the vertical force acting all round the tube.
If, now, s be the specific gravity of the liquid, h the
mean height to which it is raised or depressed, then
Digitized by VjOOQ IC
Law of Diameters. 93
we have nt^hs equal to the weight of liquid supported,
and 27rr/=7rr^^j,
ox h = —
rs
which gives the law of diameters ^ viz., that the height
of elevation or depression varies inversely with the
radius of the tube.
Since the surface-tension acts uniformly through-
out the film, it produces a resultant pressure normal
to the surface, which is always directed towards the
centre of curvature of the surface.
It follows, therefore, that in capillary elevation the
liquid surface must be concave, and in capillary de-
pression convex.
In the case of a narrow tube partly immersed in a
liquid which is in contact with air on both sides of the
tube, we see that when the surface of the liquid within
the tube is concave, the pressure immediately below the
surface is less than the atmospheric pressure by a force
due to the concavity of the surface, and that the pres-
sure goes on increasing till at a depth corresponding to
the mean level of the external hquid it is equal to the
atmospheric pressure. If the surface is convex, the
pressure immediately below the surface of the liquid in
the tube is greater than the atmospheric pressure by
a force corresponding to the difference of the mean
level of the liquid inside and outside of the tube.
Thus, in the case of a concave meniscus the resultant
normal force due to the surface-tension acts iii a
direction opposite to gravity, and so lessens its effect,
and in the convex meniscus it increases it.
Digitized by VjOOQ IC
94 Hydrostatics.
Xlli. Diffusion of Liquids,
§ 80. DiffasioiL — If two liquids of different den-
sities, and susceptible of permanent admixture, be
placed in contact with each other, they gradually
become intermixed. The process by which these
liquids combine is known as diffusion,
§ 8t. Oraliam's Experiments. — The phenomena
of diffusion were first carefully investigated by Pro-
fessor Graham, and the results of his early experiments
were published in 1850. In conducting these experi-
ments Graham employed a number of small phials,
each holding about 114 cc of water. The necks
were carefully ground and of a uniform aperture of
about 3*15 c. in diameter. Into these
'^* ^^' ^ phials he poured solutions of various salts,
so that the liquid rose as far as the shoulder
of each phial. The phials were then very
carefully filled with cold water. Thus
charged, each phial was closed by a
glass plate and then placed in a glass
jar, containing sufficient water to cover
the mouth of the phial and to rise at least
2*5 c above it. The plate was then
carefully removed and the whole apparatus main-
tained for a certain period of time at a fixed tempe-
rature. The phial was then withdrawn and the water
in the jar having been evaporated, the amount of salt
that had passed from the phial into the jar by this
process of diffusion was determined by weight.
A very simple experiment showing the diffusion
d by Google
Diffusion of Liquids, 95
of liquids is the following : — ^Take a tall cylindrical
jar and fill it to about two-thirds of its fig. 44.
height with a solution of litmus. Then^
by means of a funnel pour in very care-
fully some hydric-sulphate, so as to occupy
the lower part of the jar. If the jar be
set aside, the acid will be found, after
the lapse of two or three days, to have
diffused into the litmus solution, as shown
by the consequent red colour it will have
acquired.
§ 82. Besnlts of Graham's Experi-
ments. — By varying the experiments with '\
respect to the nature of the salt, the den-
sity and temperature of the solution, and the time
occupied in diffusion, Graham arrived at the following
results :
1. The increase in density in the diffusion product
corresponds very nearly with the proportion of salt in
the solution.
Taking four different solutions containing one,
two, three, and four parts of common salt to 100
parts of water, it was found that the quantities diffused
at the end of equal times were very nearly in the pro-
portion of I : 2 : 3 : 4, the variation from this result
not exceeding one per cent.
2. The quantity of salt diffused in equal times in-
creases with the temperature.
From an experiment with a 4 per cent, solution, it
was found that with a rise of temperature of i5°C., the
quantity of salt diffused increased somewhat more than
one-third
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96 Hydrostatics.
3. The rate of diffusion for weak solutions of the
same salt is nearly uniform, but varies considerably
with the nature of the substance diffused.
Of the several substances employed in these
experiments hydric-chloride was found to be the
most diffusible, and albumen one of the least diffusible.
Intermediate between these was common salt Com-
paring common salt with sugar-candy and albumen,
it was found that with solution of 20 parts of the solid
substance in 100 parts of water, exposed for eight
days at a temperature of 16° C, the specific gravity of
the diffused solutions were respectively i'i26 ; 1*070 ;
and 1-053.
The following table shows the approximate times
of diffusion of equal weights of different substances,
taking that of hydric-chloride as unity :
Hydric-Chloride. . . . .1
Sodic- Chloride . . . •2*33
Cane Sugar. 7
Magnesic Sulphate. ... 7
Albumen 49
Caramel 98
4. If a solution be taken of two different salts
which do not chemically combine, it is found that each
follows its own rate of diffusion. Thus the inequality
of diffusion of two different salts supplies a method
for their separation, to a certain extent, from each
other. If a mixed solution of two corresponding
salts of potash and soda be placed in the phial, the
potash salt being more diffusive than the soda salt
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Graham's Experiments, 9/
will escape into the water outside, whilst the soda salt
will be relatively concentrated in the phial.
5. It is also found, in the case of very weak solu-
tions, that a salt will diffuse into water which already
contains some other salt in solution, showing that the
diffusion of one salt is not sensibly resisted by the pre-
sence of another.
§ 2^^, CrystalloidB and Colloids. — By consider
ing the different diffusibiHty of different substances,
Graham found that all bodies might be referred
to two classes, which he called crystalloids and col-
loids. The properties of these two classes of sub-
stances are very different. Bodies susceptible ol
crystallisation belong to the former class. They are
highly sapid ; they generally form a solution which is
very slightly viscous, and they diffuse rapidly through
water, or through a porous diaphragm. Colloids, on the
other hand (so-called from KoXXiy, glue), to which class
belong starch, gum, hydrated alumina, albumen, gela-
tine, and many organic compounds, are insipid and
have very feeble chemical relations. They diffuse very
slowly in water ; but they form a medium, like water,
which arrests the passage of other colloids, but through
which crystalloid substances are capable of diffus-
ing. A peculiar property of these substances is their
mutability. They pass very easily from the liquid to
the curdled condition. They are often largely soluble
in water, but are held in solution by a very feeble
force. They never crystallise, and are generally dis-
tinguished by the sluggishness of the molecules com-
posing them.
§ 84. Dialysis. — Many insoluble colloids are per-
H
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98 Hydrostatics.
meable, as water is, to highly diffusive substances, but
effectually resist the passage through them of other
colloid substances which may be in solution. If,
therefore, a solution containing a crystalloid and col-
loid substance be placed in a vessel containing water
from which it is separated by a septum, or membrane,
formed of some insoluble colloid, the crystalloid will
pass through the septum, and the other less diffusive
substance will remain behind. This process of sepa-
ration by diffusion through a colloid septum is termed
dialysis.
Experiment, — ^Take a sheet of very thin and well-
sized paper having no porosity (so that if wetted on
one side the other side remains dry) ; let the paper be
thoroughly wetted and then laid on the surface of some
water contained in a small basin of less diameter than
the width of the paper. Having made a small de-
pression in the paper, so as to form a cavity, place
within it a mixed solution of cane sugar and gum arabic
containing about 5 per cent of each. In the course
of twenty-four hours the water below will be found to
contain about three-fourths of the sugar, which will
have passed through the paper, leaving the gum behind.
What takes place during the process of dialysis
may be thus explained : — A soluble crystalloid is ca-
pable of separating the water from the colloid, with
which it is feebly united in the septum. It thus ob-
tains a liquid medium for diffusion. The soluble
colloid, on the other hand, is unable to.separate the
liquid from the colloidal substance of the septum,
with which it is chemically combined, and is thus
unable to find a way for its own passage by diffusion.
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Dialysis — Osmose.
99
The process of dialysis is extensively employed for
separating poisonous crystalloids from organic sub-
stances with which they may be mixed ; and thus
separated they yield more easily to the methods of
chemical analysis.
§ 85. Osmose. — Intimately connected with the
phenomena of diffusion is the interchange of liquids
through porous diaphragms, which is known as osmose,
Dutrochet ^ was the first to give his attention to this
subject The endosmometer, an instrument invented
by him, illustrates this process. It consists of a long
tube, connected at its lower end with a membranous
bag which forms a reservoir. This is filled with a
solution of sugar or gum, and Fig. 45.
is kept in water. After a time
the liquid rises in the tube>
and the level of the water falls
in which the endosmometer is
placed. Moreover, traces of
the substance contained in the
bag are found in the water out-
side. This shows that an inter-
change of the liquids has taken
place, and that more liquid has
passed inwards through the mem-
brane than has escaped outwards from the reservoir.
To this flowing-in of the liquid Dutrochet gave the
name of endosmosis^ and he called the reverse process,
by which the liquid passes out from the membranous
bag, exosmosis.
From numerous experiments Graham was led to
' Nouvelles Recherches sur Tendosmose et Texosmose, 1828.
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ICX) Hydrostatics.
infer that in order to induce osmotic action between
two liquids they must each be capable of acting che-
mically, but in different degrees, on the diaphragm
separating them. Where porous materials are used
not susceptible of chemical decomposition the osmo-
tic action is very slight When either of the liquids,
by means of capillarity or by its action on the septum,
or possibly by both processes, has effected a passage
through the diaphragm, diffusion takes place, and the
liquid rises in the tube. No perfectly satisfactory ex-
planation, however, of these opposing currents of fluid
has as yet been given.
The absorption of liquids by the spongioles of
plants, and the interchange of liquids, which is con-
stantly taking place in the animal body, in the pro-
cesses of nutrition and secretion, are probably due to
osmotic action and liquid diffusion.
§ 86. Holecnlar Structure of Liquids. — ^The
phenomena of diffusion are interesting as affording us
some insight into the molecular structure of liquid
bodies. For, seeing that the molecules of one liquid
are readily displaced by the molecules of another
liquid into which they are diffused, it is evident that
these molecules must be in a state of constant agita-
tion and must be capable of continually changing
their positions relatively to one another. In this way,
therefore, the interchange of the molecules of two
liquids in contact with each other can be explained ;
and the degree of diffusibility of any soluble substance
would depend on the extent of the excursion which
the molecules of the solution undergo, or on the dis-
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Diffusion of Liquids loi
tance through which a molecule can travel before it
has its direction changed by impact upon another.
Thus the sluggishness of a liquid colloid would be
accounted for by the very small range of the motion
of its molecules \ and this explanation is supported
by the fact already referred to, that these liquids are
frequently highly viscous, and pass very readily into
the solid state — a condition of matter in which the
particles, though probably capable of revolving about
an axis and of oscillating within certain limits about
their mean positions, are not supposed to undergo any
motion of translation whatever.
d by Google
t02 Pneumatics.
CHAPTER VI.
THE PRINCIPLES OF PNEUMATICS.
XIV. General Properties of Gases. — Atmospheric
Pressure,
§ 87. Pneumatics. — The application of the prin-
ciples of dynamics to the investigation of the pheno-
mena presented by gaseous bodies constitutes a branch
of Hydrodynamics known as Pneumatics.
§ 88. Expansibility and Compressibility of
Chtses. Experiments. — The characteristic properties
of gaseous as distinguished from liquid bodies are
shown by the following experiments : —
1. If a small quantity of gas be admitted into an
empty vessel it will immediately expand, so as to oc-
cupy the whole of it.
2. If a bladder containing gas be placed under the
receiver of an air-pump, from which the air is gradu-
ally removed, the gas contained in the skin is found
to expand with the diminution of the external pres-
sure.
3. If a cylindrical vessel fitted with a piston con-
tain a certain quantity of gas, and the piston be
pressed downwards by a weight or some other force,
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A tmospheric Pressure, 1 03
the volume of the gas will be diminished ; and if the
pressing force be removed the gas will expand.
It thus appears that the volume, which a certain
quantity of gas occupies, depends on the pressure to
which it is subjected, and changes with the size of the
vessel containing it
§ 89. The Air. — ^The earth is surrounded by a
gaseous envelope reaching to a considerable height
above the earth's surface. It consists of a mixture
of two gases, nitrogen and oxygen, with small and
variable proportions of other gases, especially of car-
bonic-dioxide. By the experiments of MM. Pictet
and Cailletet, previously referred to (§ 4), air and
its two principal constituents have been separately
reduced to the liquid state. Carbonic- dioxide has
'long since been known to be a liquefiable gas.
As the majority of experiments connected with
gases must necessarily be performed in air, it is very
desirable, at starting, to consider some of its proper-
ties, as well as the various effects due to its actioa
§ 90. The Air has Weight— That the air is a
heavy fluid was not known till comparatively modem
times. Its invisibility, and its relative position with
respect to all ponderables underlying it, may have
helped to conceal this important fact from the know-
ledge of early observers. It appears that Aristotle,^
suspecting the truth of this fact, and wishing to verify
his belief, weighed a skin first empty and then inflated
with air \ and finding it to weigh the same in both cases,
concluded that au: was a weightless fluid. The failure
of this experiment was due to his having overlooked
» B.C. 384-322.
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t04
Pneumatics.
Fig. 46.
the fact that the skin when inflated with air cx:cupied
a correspondingly larger volume, and that its weight
was diminished by that of the air which it displaced,
which was exactly equal to the air which it contained.
Aristotle's experiment was regarded as decisive for
many centuries ; and it was not till the time of Gali-
leo ^ that the air was known to be a
heavy fluid. It was reserved, how-
ever, for later philosphers to see in
this fact the true cause of a variety
of phenomena which were previously
unexplained.
Otto Guericke^ is said to have
devised the following experiment for
showing that the air has weight : —
By means of the air-pump, of
which he was the inventor, he ex-
hausted a glass globe, fitted with a
stopcock, of its contained air. He
then very carefully weighed the globe, and as soon
as the scalebeam was perfectly horizontal he opened
the stopcock. The air immediately rushed in and the
globe descended. From this he very justly inferred
that the additional weight which had to be placed in the
other scalepan to restore equilibrium was equal to the
veight of the quantity of air admitted into the globe.
§ 91. Measure of Atmospheric Pressure. Torri-
cellfs Experiment. — To Torricelli is due a most im-
portant experiment, which not only shows us that the
air has weight, but also gives us a measure of its pres-
sure-intensity at any point.
• Born at Pisa 1564 ; died 1642. '^ Of Magdeburg ; 1602-1686.
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TorricelWs Experiment, 105
Take a glass tube about 34 inches in length, open
at one end and closed at the other. Fill it carefully
with mercury, and, placing the thumb over the open
end, invert the tube with this end under the yiq.^i,
surface of some mercury contained in a cup.
On removing the thumb the mercury will be
found to sink somewhat, and after a time will
remain stationary in the tube, with its surface-
level about 30 inches above the surface-level of
the mercury in the cup.
Now, since the pressure-intensity is the
same at all points in the same horizontal plane
of a liquid in equilibrium, the pressure-intensity
at any point of the external surface of the mer-
cury is equal to that along c d. But the pres-
sure-intensity on c D is equal to h s, where /i is
the difference of level of the mercury inside and out-
side the tube, and s is the weight of a unit volume of
mercury. Hence the intensity of the atmospheric
pressure on the external surface of the mercury is ^ j* ;
and the pressure on any area of the same size as the
section of the tube is equal to the weight of a column
of mercury having that sectional area for base and
the difference of level of the surface-level of the liquid
inside and outside the tube for height.
The space above the surface of the mercury in the
tube is called the Torricellian vacuum. It is really
occupied by mercury vapour, the effect of which on
the height of the column may generally be disre-
garded.
§ 92. EflFect of Atmospheric Pressure. — The
downward pressure exerted by the atmosphere is the
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io6 Pneumatics.
real cause of a number of phenomena which, at one
time, were explained on the supposition that * Nature
abhors a vacuum.'
It was observed that whenever a fluid was forcibly
removed from a vessel, air or some other fluid took its
place; and this tendency to occupy a vacant space was
foimd to be sufficiently strong to counteract gravity and
Fig. 48. other forces. Thus if the bulb of
a glass vessel, the stem of which is
under water, be slowly heated, bub-
bles of air will rise through the
water, owing to the expansion of
the air in the bulb; and if the source
of heat be afterwards removed, the volume of the air
will gradually diminish, and the water will rise in the
stem. The same effect is produced in a tube open at
both ends by ordinary suction.
From observations such as these it was inferred
that Nature abhors a vacuum ; and this general pro-
position, which expresses, though somewhat vaguely,
the results of experiments made, within certain
limits and under particular conditions, was sup-
posed for many centuries to constitute an undeniable
law of Nature. Galileo was the first to show that
this so-called law was not universally true. He
found that water would not rise in an empty tube
to a greater height than about 2,Z feet, and he very
rightly concluded that the force supporting this column
of water was the pressure of the atmosphere. Nature's
abhorrence of a vacuum was thus proved to have a
practical limit Galileo does not, however," appear
to have recognised all the consequences of his own
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Barometers,
107
discovery. It was left to Pascal, by an extension of
Torricelli's experiment, to give to this important prin-
ciple its full development. By performing Torricelli's
experiment at different elevations above the sea-level,
Pascal successfully disproved the old theory, which he
showed was the result of insufficient and circumscribed
observations ; and by clearly establishing the fact that
the column of liquid supported varied with the height
of the atmosphere above it, he proved conclusively
that the atmospheric pressure was the true cause of all
the phenomena in question.
§ 93. Barometers. — ^A barometer in its simplest form
consists of a tube such as that used in Torricelli's ex-
periment, and containing a column of liquid supported
by the atmospheric pressure which it serves to meiasure.
The height of the column is the difference in level
between the surface of the liquid in the tube and in the
cup. This height varies with the liquid employed and
the conditions of the atmosphere. fig. 49.
The great specific gravity of mercury
renders that liquid best adapted for
the construction of a barometer, as
the height of the column is corre-
spondingly small. With a mer-
cury barometer the average height
is 76 cm.
A form of instrument very com-
monly in use is that known as the
siphon barometer.
It consists of a bent tube open
at one end and closed at the other. Mercury having
been introduced into the tube in sufficient quantity to
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lo$ Pneumatics,
fill the longer leg, the tube is placed in a vertical posi-
tion when the mercury assumes a position of equili-
brium, as shown in fig. 49.
Its principle is the same as that of the instrument
already described. The pressure-intensity at d, due
to the column of mercury e d, is equal to that at c,
due to the column of air above it
In this form of instrument the cup containing mer-
cury is not required; the difference of level of the sur-
faces of the mercury in the two branches of the tube
measures the pressure due to the atmosphere.
§ 94. Barometric Corrections. — Absolute pressure
per unit area. — ^When the barometer is used for deter-
mining differences of atmospheric pressure at different
places, or at the same place at different times, certain
corrections must be made in the observed height of
the column, of which the following are the principal: —
I. Correction for Temperature, — A column of
mercury that measures 76 cm. at 0° C. is found to
have increased in length when measured at some
higher temperature, in consequence of the expansion
of the liquid under the influence of heat. It is neces-
sary, therefore, in comparing barometric heights at dif-
ferent places and at different temperatures to calculate
what the height of the column at each place would
have been if the temperature had been the same; and
it is usual to reduce the observed height of the column
in each case to the height of a column that would
produce the same pressure at 0° C. In order to ob-
tain accurate results, a further correction must be made
for the expansion of the scale on which the measure-
ments are marked.
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Barometric Corrections. 109
2. Correction for Capillarity. — ^When the mercury
IS contained in a narrow tube, the internal diameter of
which is less than three-quarters of an inch, the column
is, as we have seen, sensibly depressed in consequence
of the tension of the convex surface of the mercury.
In this case the barometric reading, as reckoned from
the top of the convex meniscus to the surface-level of
the mercury in contact with the air, indicates a pres-
sure a little less than that of the atmosphere, and the
necessary correction must be added. The error due
to capillarity is only observable in narrow tubes, and
is slightly different when the mercury is rising from
what it is when the mercury is falling in the tube.
3. Correction for Difference of Sea-level, — Owing to
the compressibility of air its density at places near the
sea-level is greater than at places higher up. The
exact law according to which the density of the air
decreases as we ascend will be considered later on ;
but even if the density of the atmosphere were uni-
form, the pressure-intensity would be greater at places
of low than of high elevation. In fact, if a barometer
is carried up a mountain it indicates a continuously
decreasing atmospheric pressure ; and, from the dif-
ference in height of the barometric column at two
places, the difference of the elevation of the two sta-
tions above the sea-level can be determined.
But the height of the barometric column is liable
to frequent fluctuations, owing to occasional and acci-
dental causes, the effect of which it is often required
to ascertain. If, with this object, observations of the
barometer are made at two stations at different heights,
an addition must be made to the observed reading at
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I TO Pneumatics.
each place, so as to reduce it to what it would be at
the sea-level. Thus, if the barometric column falls,
on the average, 60 mm. in being carried from the sea-
level to station A, and 85 mm. to station B, 60 mm.
and 85 mm. must be added to the observed reading
at each of those stations.
4. Correction for the unequal value of g. — In esti-
mating the pressure of the air on a given area, by the
weight of a column of mercury, we obtain a result
which varies with the value of ^, and affords, therefore,
no uniform standard for the comparison of the atmos-
pheric pressure in places situated in different latitudes.
At Paris, where ^= 980*94 cm., the pressure repre-
sented by a mercury column of the same height is less
than at Greenwich, where ^ = 981 -i 7 cm. If h equals
the height of the barometric column, and //the density
of the liquid employed, the measure of the pressure
of the air per unit-area x^ghd^ absolute units of force.
If we put ^ = 76 cm., and//= 13*596, the density of
mercury, then the absolute pressure at Greenwich per
square centimetre is
98117 X76X i3'596=ioi38x 10^ cos units of force.
It would be practically convenient to take the round
number 10^ units of force per square centimetre as
the standard of atmospheric pressure. This pressure
would be represented at Greenwich by 74*964 cm., 01
29*514 inches of mercury.
The standard atmospheric pressure is commonly
called * an atmosphere ' ; and a pressure twice as great
is known as ' two atmospheres,' and so on. Thus, a
pressure of 10^ units of force would be a pressure of
ten atmospheres.
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The Siphon, 1 1 1
§95. Examplei.— (I.) Requiredthe height of a water baro-
meter when the mercury barometer stands at 30 inches, the den-
sity of mercury being 13*596.
Let h be height of water barometer,
A' „ ,, mercury barometer,
and let d and d' be densities of water and mercury respectively.
Then, atmospheric pressure per unit area ^ghd^gh'd! and
.', .4 = 30 X 13-596 = 34 ft. nearly,
since ^= i and g is constant.
(2.) If a mercury barometer standing at 76 cm. be im-
mersed 4 metres below the surface of a lake, find the height of
the column.
A pressure of 4 metres of water is equivalent to a pressure
of 4 -^ 13*596 metres of mercury, t.^., 29*4 cm. nearly. Hence
the increase of pressure will be denoted by a rise of 29*4 cm.,
or the height of barometric column will be 105 '4 cm.
(3.) What is the absolute pressure, in units of force, due to a
height of 100 metres of sea-water of density i '027, g being
981?
The pressure equals 981 x 1*027 x 100 x 100= 1*0075 ^ 'o'
units of force per sq. cm., and is equal to about 10 atmo-
spheres.
J^ § 96. The Siphon. — This is an instrument the
action of which depends on the atmospheric pressure.
It is used for drawing off liquid from one vessel to
another, and consists of a bent tube with arras of
unequal length.
The siphon must be first filled with the liquid to
be drawn off, and the shorter arm being temporarily
closed, and then plunged beneath the liquid, a con-
tinuous flow will take place.
The action is thus explained : —
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TI2
Pneumatics,
Fio,
Consider era vertical element of fluid occupying
the section of the tube at its highest part.
Then, if a d be a horizontal plane
continuous with the surface of the
liquid, the pressure on the side
A c of the element c r is
^— c E,
where ^is the height of a column
of the same liquid, corresponding
to the pressure-intensity of the at-
mosphere; and the pressure on
the other side is Jf—c f.
Hence, as c f is greater than c e, there is a resultant
force acting from right to left, and causing the liquid
to flow along c d. In this way a continuous flow is
maintained. It is of course understood that c e is
not greater than H', for if c e were greater than If^
the liquid would commence to flow back along the
limb c A. Whatever cause may diminish If diminishes
the maximum height over which the liquid can be
carried. Should the surface-level of the liquid in the
vessel fall below b the direction of the resultant force
would be changed, and the flow would consequently
cease.
The resultant force acting on element c r is
the pressure represented by the column of liquid
^— CE — (^— Cf) = CF — CE = EF ;
and, hence, the velocity of efilux is ^/2g ef,
neglecting friction, and supposing the pressure at b to
equal the pressure at a.
d by Google
A tmospheric Pressure, 1 1 3
■' ^ ^ Exercises. IX.
1. If in ascending a mountain the barometer falls from
76 cm. to 51 cm. find the decrease in pressure on an area of
one square metre.
2. A mercury barometer stands at 29*5 in., and the specific
gravity of mercury is 13*6 : find the specific gravity of oil, if a
column 36 ft. 6 in. in height can be supported by the at-
mospheric pressure alone.
3. When the ordinary mercury barometer stands at 30 in.
find the whole atmospheric pressure on a surface the area of
which is 10 square feet.
4. At what depth below the surface of a lake will the baro-
nieter indicate a pressure of 50 in. , when the pressure of the
atmosphere is 30 in. P**^ *
5. In a siphon barometer of imiform bore the level of the
mercury in the open end falls through 4 mm. : what change of
pressure does this indicate ? i
6. If the sectional areas of the open and closed branches of
a siphon barometer are as 4 to i, through what distance will
the mercury move in the closed branch, if the mercury in the
ordinary barometer rises one inch ?
7. If the specific gravity of air is 0*0013 when the baro-
meter stands at 76 cm., find its sp. gr. when the barometer
stands at 58 cm.
8. Supposing the average barometric height to be 30 in.,
the sp. gr. of mercury 13-6, and of air 0*0013, ^"^^ the
height of the atmosphere, supposing the density to be uniform
throughout.
9. A barometer is observeii to fall ^ of an inch when car-
ried up 88 feet of vertical height : how much wovdd it fall if
taken 1 10 wds up a hill rising i in 3^?
10. If the specific giravify of niercury is 13*6, what ought to
be the length of a water barometer inclined to the horizon at
an angle of 60^, the mercury barometer standing at 30 ins. ?
1 1. If the height of the column in an ordinary barometer is h^
and the tube is inclined through an angle of 30® from the ver-
tical, what will be the barometric reading?
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1 14 Pneumatics,
12. A small bead of glass floats in the mercury of an ordinary
barometer : does it affect the barometric reading ?
^ 13. Find the greatest height over which water can be car-
ried by a siphon at the top of a mountain where the mercury
barometer stands at 50 cm.
14. Over what height can a liquid whose sp. gr. is j be
carried by a siphon, when the height of the mercurial barometer
is hf and sp. gr. of mercury is /?
15. A cylindrical body 40 inches high floats in water at the
ordinary atmospheric pressure with 10 inches of its height im-
mersed, the sp. gr. of air being '0013. The body, with the
vessel in which it floats, is then placed under a bell in which the
atmospheric pressure is ten times as great : what part of the body
will now be immersed ?
16. If the atmosphere press with 15 lbs. on every square
inch, and the weight of a cubic foot of water be 62 J lbs., find
the total pressure on a rectangular plane surface, placed verti-
cally with its upper edge a foot deep in water, and its lower
edge 3 ft. deep, the rectangular surface being 6 ft. long and 2 ft.
wide.
17. Find an equation for determining the internal radius of
a globe of thickness / and specific gravity f, which will just float
in air of specific gravity j, when filled with gas the density of
which as compared with air is d.
18. Taking the pressure of the atmosphere at 15 lbs. per
square inch, the height of the salt-water barometer at 30 feet,
calculate the pressure-intensity at 50 fathoms depth in tons per
square foot. If this pressure acted for a second on a square yard
of the stem of a vessel weighing 300 tons, what velocity would
it communicate, there being no resistance to the motion ?
XV. Relation betiueen the pressure and volume of a
gas^ the temperature remaining constant. —
Boyl^s Law,
We have seen that the characteristic quality of a
gas is its expansibility and compressibility. We come
# Digitized by VjOOQIC
Boyle's Law,
IIS
FiG» SI,
now to consider the quantitive relation that exists be-
tween change of volume and change of pressure, the
temperature remaining constant. This relation can
be inferred from a series of experiments, of which
the following are instances : —
§ 97. Experiments. — For pressures greater than
the atmospheric pressi/re, — i. A simple form of ap-
paratus for these experiments consists of a uniform
bent tube, having one branch open at the top and
considerably longer than the other, which
is closed. The shorter branch is some-
times fitted with a screw-cap so that the
pressure of the enclosed gas may be more
easily regulated. The tube itself is gra-
duated, or a scale is fixed to each branch.
If the cap, which should fit au:-tight,
be first unscrewed, and some mercury
poured into the tube, the mercury will
rise to the same level in both branches.
If now the cap be screwed on again some
air will be enclosed under a pressure equal
to that of the external atmosphere. Call
this pressure 76 cm., and suppose the mercury to
stand at o in each branch. Pour more mercury into
the tube, and the level of the mercury will be lower
in the shorter than in the longer branch. The volume
of the air confined in the shorter branch is now
diminished, and its pressure is increased by that of
the column c d.
Suppose the air in the tube to have originally
occupied 20 divisions of the tube, or 20 cm., and that
mercury has been poured into the tube to a height of
Digitized by VjOOQ IC
o
ti6
Pneumatics,
23 cm. above its former level in the longer branch,
then the mercury in the shorter branch will stand at
4 cm. above o, and, consequently, the enclosed air
will occupy only 16 cm. The increase of pressure, due
to the difference of level, is therefore 23 — 4 = 19 cm.
Now, if we examine these numbers we shall find
20 : id:: 19+76 : 76
i^e. ::95:76,
or, 20x76= 16x95,
which shows that the variation in the volume is in*
versely as the variation in the pressure ; or, what is
the same thing, that the product of the volume into
the pressure is constant.
V
H
h
H-h
H-h-^n-g-P
PV
150
34-1
7*3
26-8
1027
1540
120
624
lO'O
52-4
1283
1539
10 -o
90-I
II-8
78-3
154-2
1542
V
io6-o
127
93-3
169*2
1539
8-3
I2I-3
I3'3
loS-o
1839
1526
7*4
I47-0
14-1
132*9
208-8
1545
67
i67'5
148
1527
228-6
1547
6-4
1802
150
165-2
241-1
1537
Mean value . .
1539-3
F«= volume of air In cu
b. cents. ; H^ height in c
ents. of
merc\iry in open lira
b ; h — height in closed lim
b; 759
a» height of baromet
er.
On comparing the results of a number of actual
experiments, the value of the product of the volume
into the pressure will be found to vary slightly, the
extent of the variation depending on the degree of
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Boyl^s Law,
\\^
accuracy with which the experiments are performed.
In the above table are found the results of a few
experiments in which the measurements were roughly
made, the fractional parts of a centimetre being esti-
mated by eye. They serve, however, to show how
nearly constant is the value of P V.
2. For pressures less than the atmospheric pressure.
To determine the relation between the change in
the volume and pressure of a gas, when the gas
expands, we take a barometer tube fitted with a
screw-cap, and having opened the tube, partly im-
merse it in a vessel containing mercury. We then
close the tube, by means of the screw, and thereby
enclose a certain quantity of air at the p^^
ordinary atmospheric pressure. 'J*he tube is
now raised vertically upwards, and the en-
closed air expands, its pressure being di-
minished by that due to the difference of
surface-level of the mercury in the tube and
in the vessel. Thus, if ^ ^ = Fbe the volume
occupied by the air when the mercury
stands at the same level in the tube and in
the vessel — that is, under the ordinary atmos-
pheric pressure H^ and if a //= V be the
volume occupied at the pressure H-cd^=^P\
then, on reference to the scale, it will be
found that
ah : ad:: H—cd: H,
or, VH^ rP'.
§ 98. Statement of Boyle's* law.— From experi-
* Robert Boyle born at Lismore in Ireland, 1626 ; died 1691.
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nen- '
Il8 Pneumatics,
ments such as these performed with different gases,
and at different temperatures, a law has been estab-
lished, which we shall see is not accurately true for any
gas, but approximates very nearly to the truth for
those gases which are not easily reducible to the
liquid state. The law is known as Boyle's law or Mar-
riotte's law, as these two philosophers are said to
have arrived independently and by similar experiments
at the same result. It may be enunciated thus : —
The volume of a gds varies inversely with its pres-
sure, when the temperature remains constant.
Thus, if the volume V change to V\ while the
pressure changes from/ to/', we have : —
V : v'::p' \p) or Vp=-vy,
As the mass of the gas remains the same in the
experiments already described, it follows that the
density of the gas must increase as the volume
diminishes, and vice versL Hence Boyle's law may
be stated thus : —
The pressure of a gas is proportional to its density,
tJu temperature remaining constant.
For a perfect gas, therefore, we have the following
relations : —
V' p d '
Another statement of this law, due to Professor
Rankine, places the law in a very clear light : —
* If we take a closed and exhausted vessel and
introduce into it one grain of air, this air will, as we
know, exert a certain pressure on every square inch of
the surface of the vessel. If we now introduce a
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Dalton's L aw, 1 1 9
second grain of air then this second grain will exert
exactly the same pressure on the sides of the vessel
that it would have exerted if the first grain had not
been there before it. Hence we may state, as the
property of a perfect gas, that any portion of it exerts
the same pressure against the sides of a vessel as if
the other portions had not been there.' *
§ 99. Dalton's Law.^ — This is an extension of
Boyle's law for a mixture of different gases. If several
different gases, which do not act chemically on one
another, are placed in a vessel, the pressure on the
sides of the vessel is the sum of the pressures due to
the different gases. Thus, suppose each of the gases,
if in the vessel by itself, to exert pressures the intensi-
ties of which are/i, ^2, /a, &c., respectively, the inten-
sity of the whole pressure exerted by the mixture is
P\ -^ Pz 4-/3 + &c. Rankine's statement of Boyle's
law shows this to be the case for different parts of
the same gas ; hence Boyle's law may be considered
as a particular case of Dalton's,^ which may be thus
stated : — When a mixture of several gases ^ at the same
temperature^ is contained in a vessel, each produces the
same pressure c^ if the others were not present,
§ 100. Examples.— (I.) In a bent tube, open at one end
and closed at the other, the mercury stands at the same level in
both branches, and the contained air occupies 30 cm. at the
normal atmospheric pressure — viz. 76 cm. If the section of
the tube is 10 sq. cm., what volume of mercury must be poured
into the tube to compress the air to 20 cm. ?
If / be the pressure of the air when occup3nng 20 cm., wc
C. Maxwell, Theory of Heat, pp. 27-8. « 1801.
■ Bom in Cumberland 1766 ; died 1844.
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I20 Pneumatics,
have by Boyle's law 20 x /« 30 x 76, .'.;?= 114, of which 76
is due to atmospheric pressure. Hence the difference of level in
the two branches equals 38 cm. ; and as the mercury has risen
10 cm. in the closed branch the quantity of mercury introduced
equals (20 + 38) ID = 580 cubic cm.
(2.) The mercury in an ordinary barometer stands at 30 in.,
and the sectional area of the tube is one square inch. A cubic
inch of air is admitted through the mercury into the vacuum
above, and depresses the column through 4 inches : find the
size of the vacuum. • ^
When the cubic inch of air is admitted into the vacuum it
first fills the vacuum, and then with its diminished elasticity,
consequent on its expansion, it depresses the mercury until the
pressure of the column of mercury, together with the elasticity
of the air, equals the atmospheric pressure.
Suppose the vacuum to measure x inches ; then the air
which under a pressure of 30 inches occupied i inch is found to
occupy ;c + 4 inches under a pressure of 30— (30— 4) = 4 in.
Hence, by Boyle's law, 4 x (^ + 4) = 30, or ;c=3'5 in.
§ loi. Graphic representation of Boyle's Law. —
Take o x, o^, two straight lines at right angles to each
*'^G. 53. other, and along ox
mark off o a, to repre-
sent the number of
units {V) of volume
which a certain quan-
_n tity of gas occupies at a
-^ pressure/. Then, if we
mark off o b, o c. . . to
represent 2 ^ 3 F .... we know by Boyle's law that
the pressures corresponding to these volumes will be
^, ^ .... Hence, if a p be drawn vertically from
23*
A to represent /, then' the lines b q = — , c r =
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\
K
Graphic Representation of Boyle's Law. 1 2 1
— , &c. will represent respectively ^^ ?-, &c., and the
3 . ^. ^
curve drawn through these points will serve as a gra-
phic representation of Boyle's law.
Now, it is evident that no amount of compression
can reduce the volume of a gas to zero, and no
amount of expansion can wholly destroy the pres-
sure which a perfect gas exerts ; it follows, therefore,
that the curve will approach on either side nearer and
nearer to the lines o x and oy, but will never meet
them.
Since, also, Vp = V'p' = V"p", we see that the
product of the volume and the pressure remains con-
stant, whilst the separate factors vary, and therefore
the curve p q r is such that if p be any point in it,
the area of the rectangle o a p a is constant for all
positions of p.
Now, if X represent any volume measured along
ox, and J' the corresponding pressure measured ver-
tically upwards, we have xy ^=z o. constant, and the
curve corresponding to this equation is known as the
rectangular hyperbola.
§ 102. Graphic representation of the work done
in changing the volume of a gas from Fto V'^ the
temperature remaining constant.
Let a volume V of gas be en- fig. 54.
closed in the cylindrical vessel d b a c A yy c
by the piston m n, and let the pis-
ton be moved through m m' so that ^ ^ w i>
the volume becomes V, Then if p be the pressure
of the gas, when the volume is V^ the work done
in changing the volume to V would be / x m m',
Digitized by VjOOQ IC
122 Pneumatics.
supposing / to remain the same throughout. But
this is not the case, for the pressure increases as the
volume diminishes. If, therefore,/' is the pressure
of the gas when its volume becomes F', the work
done is equal to some quantity the value of which
Ues between / x m m' and />' x m m .
Now, suppose o a (fig. 55) represent the volume
V^ and o a' the volume F', a p, a' p' the correspond-
F»G. 55. ing pressures ; then, if we draw
the lines p q, p' q', parallel to
o A, the work done is repre-
sented by the area of a figure
which is greater than the rect-
angle p Q a' A, and less than
the rectangle p' a' a q'. Now,
by reasoning similar to that employed for finding the
space described when a body moves with an increasing
velocity, it can be shown that the area of the figure
I p' a' A represents the work done.
§ 103. Limits of Boyle's Law. — By experiments
conducted by Regnault ^ and by Despretz* it has been
observed that Boyle's law is not perfectly true for any
actual gas. For air and all gases that do not readily
liquefy under pressure the law is found to be a very
close approximation to the results of experiments ;
but for easily liquefiable gases, such as carbonic di-
oxide and ammonia, the volume is found to decrease
more rapidly than the pressure increases ; and the
divergence of the law from the true results is greater
as the gases approach their point of liquefaction.
» 1827, 1847. * 1827.
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Limits of Boyle's Law.
123
Thus carbonic-dioxide under a pressure of twenty
atmospheres is found to occupy only four-fifths of the
volume given by the law.
Regnault devised very careful experiments for
showing the exact relation between the pressure and
volume of different gases and the extent of the diver-
gence of this relation from Boyle's law. He arrived
at the following result : that whereas, according
to Boyle's law, Vj> = P/, or -/, -1 = 0, the ex-
Vp
Vp'
pression -^J--, — i is a quantity having a small posi-
tive value for all gases except hydrogen, and increases
gradually with the pressure.
If the distances o a, o b, o c, measured along the
horizontal line o x represent pressures of one, two,
Fig. 56.
and three atmospheres, and if the vertical lines a a,
B b, c c represent the values of ^^ — i for these
pressures, the curve formed by joining o « ^ ^ . . . is a
graphic representation of the divergence of the results
of direct experiment from Boyle's law.
Seeing that f^ — i is a positive quantity, Vp
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124 Pneumatics.
must be greater than V p', and therefore the value of
V must be less than it would be if Vp = V' p\ in
accordance with Boyle's law. Hence it appears that
gases are more compressible than is in accordance
with Boyle's law.
In this respect, however, hydrogen differs from
other gases, and this exception shows that the law is
more complicated than it would seem to be even with
the extension above given. It has, however, been
proved that the true relation of the volume of a gas to
its pressure depends to some extent on the tempera-
ture. The curve exhibited in fig. 56 is not the same for
all temperatures. For nearly all gases the value of
V V
the expression -j^ril, — i contmually decreases as the
V p
temperature rises, and we are 'thus led to expect
that if the temperature were sufficiently high this
quantity would pass through zero and become nega-
tive ; or the curve, after coinciding with the line o Xy
would reappear on the other side. It should seem,
tlierefore, that at a certain temperature varying with
each gas Boyle's law is strictly accurate, and that
for higher temperatures the law of the divergence
is changed, so that the density increases less rapidly
than the pressure. Now, as hydrogen which has
already been reduced to the solid state is commonly
supposed to be the vapour of a metal, it is relatively at
a very high degree of rarefaction, and this fact may be
the reason why — ^ - i has a negative value for this
gas.
§ 104. Belative Densities of the Air at di£Eerent
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Variation in Density of Atmosphere, 125
Heights. — ^We are now in a position to determine the
law according to which the density of the air changes
as we ascend from the level of the sea. The decrease
in the density of air is owing to its compressibility; but
even if the air were as incompressible as water, and the
atmosphere were homogeneous throughout, the baro-
metric column would be found to fall, in rising from
places of lower to places of higher elevation, in conse-
quence of the diminution in the height of the column
of air. In order to obtain an approximate relation
between the densities of the air at two different heights
we must neglect the accidental differences of pressure
caused by differences of temperature and moisture,
and by the altered value of the force of gravitation.
Take a vertical column of the atmosphere and sup-
pose it divided by horizontal planes Fig. 57*
into a number of strata so thin that
the density for each layer may be con-
sidered uniform and equal to that at its
lower surface. Let the height of the
column be z^ and let n be the number of _
strata, so that the thickness of each layer is -
n
let the section of the column be the unit of area.
Let d and p be the density and pressure of the
atmosphere at the surface of the earth, and let //j, d<i^ d^
• • 'iP\ip2iPz' • • • be the densities and pressures
at the successive levels.
Then, since the difference between the pressures
at the upper and lower surface of a layer must be
equal to the weight of that layer of air,
p-px = weight of lowest layer = dg^ (§ 17):
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126 Pneumatics
And since by Boyle's law the pre
density, we have -^ = -^^ -^ =
a dx d^
k {d -- dx)^a
whence ^/i = (i — S.
\ k
similarly ^, = fi^^t
\ kfi
/>., the ratio of the densities for t^
is constant, since g, k^ z and n are
fore the quantities d^ d^^ d^, d^ . .
cal progression, the heights of tl
arithmetical progression.
If, therefore, dn be the density
d \ kn)
% 105. To find the difference
stations by means of a baromet
temperature and force of gravitj
Let H and h represent the ba
at the two stations, the vertical dist
being z^
ThenIf:A::d: a
If V k
y Google
Difference of Height and Pressure. 1 27
where d\% the density of the air at the lower, and d^ the
density at the upper station. Now, \in increase without
limit, it is proved in works on Algebra, that the value
of the right-hand side of this equation becomes e
Hence ^ = ^ > and \og.S-i ] == f * ^
Or,0=^(log.e^-log.e>^)
Exercises. X.
1. A gas occupies 100 litres when the barometer stands at
76 cm, : find the increase in the volume of the gas, if the
pressure becomes 73 cm.
2. A tube 2 feet long is filled with water and inverted in a
vessel of water, with its open end below the surface. Air at
a pressure of 30 ins. is then admitted into the tube till the level
of water in the tube is the same as that outside, and the air
occupies 12 inches. The tube is now raised till the air occupies
15 inches : find the pressure of the contained air.
3. Into the vacuum above a common barometer, which stands
at 30 inches, 2 cubic inches of air are admitted, which depresses
the mercury 6 inches : if the section of the tube is one square
inch, find the size of the vacuum.
4. A vessel of 3 cubic feet capacity containing air at two
atmospheres' pressure is put into communication with a vessel of
18 cubic feet capacity containing air at \ of the atmospheric
pressure : what is now the pressure of the air in the two
vessels ?
5. A horizontal cylinder containing air is fitted with a piston
which is 10 inches from the closed end when the pressure is
15 lbs. on the square inch. If the area of the piston is 8 square
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1 28 Pneumatics,
inches, find the force that must be exerted to hold the piston at
a distance of 12 inches from the closed end.
6. If the water barometer stand at 33 feet, to what depth
must a small cylindrical vessel be sunk to reduce the volume of
the contained air to one-third of its original volimie, the height
of the vessel itself being neglected ?
7. A cylinder contains air at the ordinary atmospheric pres-
sure, and is fitted with a piston (area 10 square inches), which is
I ft. from the closed end. If the cylinder be set vertically with
its open end upwards, how far will the piston descend, the
atmospheric pressure being 15 lbs. on the square inch, and the
piston weighing 10 lbs. ?
8. Suppose the cylinder is held with the open end down-
wards, how far will the piston fall ?
9. Find what weight must be hung to the piston in the last
question to draw it down 2 inches from its original position.
10. The air contained in a cubical vessel the edge of which is
one foot, is compressed info a cubical vessel, the edge of which
is one inch : compare the pressures on the side of each vessel.
11. Forty c.c. of air are enclosed in a tube over mercury,
the height of the mercury in the tube above the level in the
vessel outside being 50 cm. (r ^«5o, fig. 52). The tube is de-
pressed until cd='}fi cm. What is now the volume of the air,
the height of the barometer being 76 cm. ?
12. A bent tube (fig. 51) has a uniform section of I square
inch and is graduated in inches ; 6 cubic inches of air are en-
closed in the shorter branch, when the mercury is at the same
level in both branches. What volume of mercury must be poured
into the longer branch in order to compress the air into 2 inches ?
The barometer stands at 30 ins.
13. The air enclosed in the shorter branch of a similar
bent tube occupies 11*3 cc, and the difference of level in the
two branches is 60*2 cm. If the barometer stands at 75*9 cm.,
find what volume the air would occupy under the atmospheric
pressure only.
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Experiments oft Diffusion of Gases, 129
XVI. Diffusion of Gases.
§ 106. DiflFtision of Oases. Experiments. — We
have seen that if two layers of different liquids are in
contact with each other, or are separated by a porous
diaphragm, a mixture of the liquids takes place, the
one diffusing into the other. Now, the same pheno-
menon is observable with gases ; but the laws of
gaseous diffusion are less complex than those of liquid
diffusion, in consequence, probably, of the greater
structural simplicity of gaseous bodies.
Fill two jars with two different gases — for example^
with chlorine and hydrogen — and let the jars be
connected by a long tube, that containing the hydrogen
or lighter gas being placed uppermost In a few
hours the chlorine will find its way into the upper jar,
as may be seen by its green colour; and the hydrogen
will take its place. Each of the jars ^^^ ^
will be found to contain the same pro-
portion of the two gases, and the gases
will remain permanently mixed. This
intermixture .takes place between any
gases or vapours which do not act
chemically on one another.
If a vessel containing nitrogen be
covered with a porous diaphragm of
some colloid substance, and be placed
under a bell-jar containing hydrogen, /lIU^
diffusion will take place, and after a
time the membrane will have become convex, showing
that the hydrogen has been passing inwards more
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1 30 Pfieumatics,
rapidly than the nitrogen has passed outwards. If the
position of the gases be reversed the contrary result
will be found. If the diaphragm be moist, and one
of the gases is soluble, its rate of diffusion is very
much increased. Thus, if a moist thin bladder be dis-
tended with air and placed in a bell-jar containing
carbonic dioxide, this gas, owing to its solubility,
passes much more rapidly into the bladder than the
air escapes from it, and very frequently breaks it,
though the rate of diffusion of carbonic dioxide into
air is really less than that of air into the gas.
§ 107. Rate of DiflFasion. — If we take a long gra-
duated tube, open at both ends, and close one end
by a plug of porous clay, and immerse the open end
in a vessel containing mercury, or water coloured, for
the sake of greater distinctness, the level of the liquid
in the tube and in the vessel will be the same,
showing that the gas must be entering the tube through
the porous plug at the same rate as
It escapes mto the outer atmosphere.
If now we bring an inverted beaker
filled with coal-gas over the closed
end of the tube, we find that there is
a bubbling of gas through the liquid,
clearly showing that the coal-gas is
entering the tube more rapidly than
the air is escaping from it. As soon
as we withdraw the beaker the liquid
rises in the tube, the pressure of the
7 mixed gas in the tube being less
than that of the atmosphere out-
side ; and if we replace the beaker
Digitized by VjOOQ IC
Graham's Law. 1 3 1
the bubbling of the gas through the liquid recom-
mences. This experiment enables us to observe the
difference only between the amount of coal-gas that
goes in and of air that escapes from the tube. We
can vary this experiment by filling the tube with other
gases than air. Suppose the tube to be first filled
with hydrogen and then inverted in the liquid, and
held in such a position that the level of the liquid
within and outside the tube is the same, the beaker
being altogether removed. After a time the liquid
will be found to rise in the tube, showing that the
pressure within the tube has become diminished, and
that the hydrogen must be passing from the tube into
the external air more rapidly than the air is entering
the tube.
It is easy to see that, if the tube be graduated,
careful experiments will show the volumes of different
gases that diffuse into the air in the same time. If
the beaker be filled with a different gas, and held over
the tube, the rates of diffusion of different gases into
each other can be ascertained.
By experiments such as these, Graham established
the law that the rates of diffusion of two gases into each
other are in the inverse ratio of the square-roots of their
densities. For instance, taking the density of air as
unity ^ that of hydrogen is 0*0692 ; and the square-
roots of these numbers being i and 0*2632 respectively,
the law tells us that the rate of diffusion of hydrogen
is to the rate of diffusion of air
as I : I-^o•2632, i.e. \\i \ 37994;
and actual experiment shows that whilst one measure
K 2
Digitized by VjOOQ IC
132
Pneumatics,
of air passes into the tube containing hydrogen, 3*83
measures of hydrogen escape into the air.
This important law can be . further verified by
taking a vessel consisting of two large receivers filled
with different gases, and connected by a tube with a
stop-cock. If now the gases be allowed to diffuse into
each other for a certain period of time, the contents
of the receivers can be analysed, and the proportion
of the two gases in each can be quantitively deter-
mined. By varying the time, and tabulating the
results, the accuracy of the law may be verified for
any two gases.
The following table shows the results of some of
Graham's experiments : —
Gas
Density
Square
root of
density
z
Rate of
diffusion
VDensity
Hydrogen ....
Marsh-gas ....
Carbonic Oxide . .
Nitrogen ....
Oxygen ....
Nitrous oxide . . .
0*06926
0-559
0*9678
0*9713
1*1056
1*527
0*2632
0*7476
0*9837
0*9856
I 0515
1*0914
37994
1-3375
1*0165
1*0147
0*9510
0*8092
3-83
1*34
I 0149
1*0143
0*9487
082
If d and d' be the densities of two gases, and D
and D' the volume of each which diffuses into the
other in the same time, then, according to the law,
D:D'::^A=:i ^^,
I
\/ d s/d
OxD^d^D'^d'.
§ 108. Kinetic Theory of Oases. — In order to ex-
plain the diffusion of gases we must suppose that the
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Kinetic Theory of Gases, 133
particles of a gas axe independent of one another, that
they are constantly moving in all directions, and with a
very great velocity. These particles during their motion
frequently impinge on one another, and the direc-
tion of their motion is consequently changed. "When
they come into contact with the sides of the vessel
containing the gas their momentum is resisted ; and
it is to this shower of particles moving with a consi-
derable velocity that the pressure of a gas is supposed
to be due. The notion that the particles of a gas
are in rapid motion, and that it is by their impact
that gases press on one another, is a very old one. It
was pointed out, not long after Newton's time, by
Daniel Bemouilli ^ ; Lesage and Prevost of Geneva
made several applications of the theory, and it was
afterwards revived in this country by Herepath. In
1848 Dr. Joule showed how the pressure of gases
might be explained by the impact of their molecules,
and he calculated the exact relation that exists be-
tween the observed pressure of a gas and the velocity
of its particles. It is, however, to Professors Clausius
and Clerk-Maxwell that this theory of the molecular
structure of gases owes its chief development.
By a method similar to that indicated in Wor-
meirs * Thermodynamics/ § 71, Joule showed that if a
vessel contain hydrogen at the ordinary pressure and at
0° C, the velocity of the particles must be about 6,055
feet per second. Now, although the velocity of these
particles is so considerable, the number of particles
occupying a given volume, say a cubic inch, is so enor-
* Born at Groningen 1700 ; died 1782.
Digitized by VjOOQ IC
134 Pneumatics.
mous that the particles move through a very small
space, and are unable to travel from side to side of the
vessel containing them, without encountering a series
of successive impacts with other particles. It follows
from this, that if the average velocity of the particles
remains the same, the pressure exerted at any point
of the vessel containing the gas depends on the num-'
ber of particles that impinge, in a given time, on the
element of area containing that point But this, of
course, depends on the number of particles contained
in the vessel, />., on the density of the gas ; for if
the volume of the gas be doubled, the average velocity
of the particles remaining the same, the number of
particles traversing a given area will be halved. Now,
the density of a gas is the ratio iDf its mass to its
volume, and hence it follows that the pressure a gas
exerts varies inversely with its volume, if its mass or
the number of particles in a given volume remain the
same; and this result is the same as that previously
obtained by experiment, and known as Boyle's law.
The notion that a gas consists of a series of parti-
cles flying about in all directions is the basis of what
is called the kinetic theory of gases. This theory,
besides explaining the phenomena of diffusion and
Boyle's law, accounts for many other facts connected
with the action of gases at different temperatures; but
this subject is beyond the range of the present volume.
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Diving Bell. 135
CHAPTER VII.
PNEUMATIC INSTRUMENTS.
XVII. Diving Bell, Pressure Gauges,
The two principles which we have now established
— viz., that the air is a heavy elastic fluid, and that the
density of a gas varies with its pressure — serve to ex-
plain most phenomena exhibited by gases at constant
temperature, and enable us to understand the action
of a great variety of pneumatic instruments.
§ 109. The Diving Bell.— If we take an ordinary
glass tumbler and immerse it vertically, with its mouth
downwards, in a tub of water, we shall find that
the enclosed air will prevent the water from rising in
the inside except to a very small height, which will
vary with the depth to which the tumbler is immersed.
The action of the diving-bell is similar. It consists of
a hollow vessel, nearly cylindrical in form, and open
at its lower end. When lowered vertically into
water the enclosed air is compressed by the weight of
water above it, and the water rises in the bell to a
height which increases with the depth of the bell from
the surface of the water. The bell is let down by a
chain; and in order that men may be able to work
in the inside, air is introduced by means of a pump
through an opening in the top, and the pressure of the
air prevents the water from rising in the bell.
Digitized by VjOOQ IC
136 Pnemnatics,
§ no. Problems on the Diving Bell.— (i.) To find
the h'iight to which water rises in the bell, at a given
depth, when no additional air is introduced : —
Let B c = 0, be the
depth of the top of the
bell below the surface of
the water. Let b be the
height of the bell, and H
the atmospheric pressure
at the surface of the water
measured by a water baro-
meter.
Then, if c a = x^ the
part of the bell occupied by the air at the depths, and
if we suppose the bell to be of uniform area inside, the
pressure on the air within the bell is equivalent to the
weight of a column of water the height of which is
Zr+ BA = ^+^ + ^. Hence, by Boyle's law :—
X H
FIO.
6<>.
Ma^^
WeM
3£r:^
1^
i^i^
^^^
A
£=£^
rrrr.. .
1^^
'■fzzi
b If+z+x
or, x^ + X (If + z) = Hb
a quadratic equation, the positive solution of which
gives the height required.
(2.) To find the volume of air at the ordinary
atmospheric pressure that must be introduced into
the bell at a given depth, to prevent any water from
entering: —
Let z be the depth of the top of the bell, and b the
height of the bell, as before. Then, if V be the volume
of the air in the bell at the normal pressure H, and F'
the volume which the compressed air in the bell at
depth z + b would occupy at pressure H, it follows
Digitized by VjOOQ IC
Manometers, 137
from Boyle's law that
V H
i\e. the volume of air introduced, at the ordinary
atmospheric pressure, is T, . V.
H
§ III. Manometers, or Pressure-Chkuges. — Mano-
meters are instruments for measuring the elastic force
of a gas contained in a closed vessel. fig. 6x.
The simplest form of manometer con-
sists of a long narrow open tube which dips
into a strong box containing mercury.
The gas, the pressure of which is to be
measured, is admitted into the box through
an opening, a, and if the elasticity of the
gas is equal to that of the air the level of ^
the mercury in the tube will be the same |
as that in the box. If, however, the L
elasticity of the gas is greater, the mercury will be
forced up the tube, and the excess of the pressure of
the gas over that of the atmosphere can be measured
by the height of the mercury in the tube.
This form of manometer cannot be used for very
great pressures, as the length of the tube renders it
inconvenient. Thus, for a pressure of two atmo-
spheres the tube must be 30 inches, and the length of
the tube must be increased 30 inches for every addi-
tional pressure of one atmosphere that is to be
measured.
d by Google
138 Pneumatics,
§ 112. Compressed Air Manometer. — For mea-
suring greater pressures this form of instrument is
better suited.
^^^' ^ _ It consists of a bent tube, one
branch of which is closed and
contains air at the ordinary atmo-
spheric pressure. This air is shut
off from the other branch by some
mercury which occupies the
lower part of the tube. The open
branch communicates with the
vessel containing the gas, the pressure of which is to
be measured. The closed branch of the tube is
fiunished with a scale.
If the mercury stands at the same level in both
branches of the tube the pressure of the gas will equal
that of the atmosphere. But if a gas or vapour of
greater pressure be admitted through g the level of
the mercury will fall in d and rise in c above the ori-
ginal level, A B.
The measure of the pressure of the gas in d is that
of the compressed air in c, together with that indi-
cated by the difference of level of the mercury in the
two branches of the tube.
In order to graduate the scale we must find the
distance a e or rise of the mercury conesponding
to a pressure of n atmospheres in d f. Let a c, the
space originally occupied by the air, equal «, and let
AE =^.
Let H be the height of the mercury in the baro-
meter at the atmospheric prebsuie, and/ the pressure
of the air in c e.
Digitized by VjOOQ IC
Pressure-gauges. 139
Then/xcE = ^x ac, or 4 = :i5 = _^ and
the pressure of the gas in df=/ + 2Ea = n If.
a—x
/. 2x'^^ (nH-\-2d)x + {n—i)Ha=^o-y
and, by the solution of this quadratic, the value of x^
corresponding to any number of atmospheres, can be
determined and the scale graduated.
§ 113. The Siphon Gauge.— For measuring small
pressures this instrument is sometimes employed. It
consists of a bent tube, open at both ends, one of
which communicates with the vessel containing the
gas. The liquid used is water or mercury.
If the gas is admitted through b, and the liquid
assumes a difference of level, p d, then the pressure of
the gas equals the atmospheric pres- fig. 63.
sure + the weight of the liquid in p d.
If, however, the liquid rises in the
other branch of the tube, then the
pressure of the gas = the pressure of
the atmosphere, minus that due to the
difference of level.
Thus, if the liquid fall through x
inches in one branch, it will rise through
x inches in the other ; and the difference of level will
be 2 X. If, therefore, a be the sectional area of the
tube and s the specific gravity of the fluid —
The pressure of the gas = atmospheric pressure
± 2XS, according as the level sinks in b or a.
d by Google
140 Pneumatics,
Exercises. XI.
1. To what depth must the top of a diving-bell 8 ft. high be
immersed under water that the air may be compressed to half its
volume, the height of the water barometer {H) being equal to
34 ft.?
2. What additional volume of air at the ordinary pressure
(^=34 ft.) must be admitted into a bell 8 feet high, the internal
section of which is uniform and equal to 20 sq. ft., and the top
60 feet below the surface, to completely fill it ?
3. A cylindrical bell, the height of which is 6 feet, is fur-
nished with a barometer that stands at 30 in., and is lowered into
water till the barometer stands at 40 in. : find the depth of the
top of the bell below the surface of the water ; sp. gr. of
mercury = 13*6.
4. A barometer marking 30 in. is carried down in a diving-
bell which is kept constantly full of air : find the depth of the
top of the bell from the surface of the water, when the barome-
ter marks 42 in., the height of the bell being 8 feet.
5. A gas, the pressure of which is 10 lbs. on a square inch,
communicates with a siphon gauge (fig. 63), the section of which
is I square inch : find the difference of level, supposing the in-
strument contains mercury and the ordinary atmospheric pres-
sure is 15 lbs. on the square inch.
6. If in the open manometer the distance of the level of the
mercury from the top of the box is </, and a gas be admitted that
depresses the mercury b within the box, find the height of the
mercury in the tube above the original level, the atmospheric
pressure being h, .
7. A cylindrical bell, 4 feet deep, whose interior volume is
20 cubic feet, is lowered into water until its top is 14 feet below
the surface of the water, and air is forced into it until it is
three-quarters full. What volume would the air occupy under
the atmospheric pressure, the water barometer being at 34 feet ?
8. Find the depth to which a cylindrical diving-bell 8 feet
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Pumps,
141
high must be sunk in water in order that the water may rise in
it 3 ft. (^= 34 ft.)
9. If the weight of a cylindrical bell is 2,000 kils. and the
specific gravity of the material 8, find the tension in the chain
when the bell is immersed to such a depth that the pressure of
the enclosed air equals 3 atmospheres, the interior volume being
I '5 cubic meters.
Fig. 64.
c
P
XVIII. Air-Pumps.
§ 1 14. Essential Parts of a Pnmp.— -A pump is an
instrument for removing a fluid from a reservoir, or
vessel containing it, by the forcible withdrawal of the
atmospheric pressure.
It consists essentially of (i) a barrel or cylinder^ c,
through which the fluid escapes ; (2) a disc ox piston ^
p, capable of moving up
and down the cylinder, into
which it exactly fits, and
worked by a handle or rody
r, attached to it; (3) a
pipe^ E, that communicates
with the reservoir or vessel
from which the fluid is to
be removed ; and (4) valves
or small apertures, with movable covers, opening one
way only, v, v', which serve to admit the fluid from
one part of the instrument to another, and to prevent
it from returning. A valve is generally found at the
end of the pipe, where it communicates with the
cylinder, and very frequently in the piston itself.
§ 115. The Air-Pnmp. — ^This is an instrument for
removing air from a closed vessel.
Digitized by VjOOQ IC
^
142
Pneumatics.
Pio. 65.
It consists of a receiver^ r, from which the air is to
be removed ; of a cylinder or barrel, a b, furnished
with a piston and
valves; and of a pipe
which serves to con-
nect the receiver with
the cylinder.
The action of the
pump may be thus
explained: —
Suppose the pis-
ton at B, when the receiver, r, is full of air. As the
piston is raised the valve m, which opens upwards, re-
mains closed in consequence of the pressure of the
external air, and the air from r rushes through the
pipe E, opens the valve n, and occupies the space
between the piston and the bottom of the cylin-
der. When the piston is first pushed down the air in
the cylinder is compressed, the valve m remaining
closed j but as soon as the pressure of the air in the
cylinder begins to exceed that of the air outside the
valve M is opened, and the enclosed air escapes as the
piston descends through it. The piston being again
raised, the same process is repeated. It should be
noted that it is only when the piston is raised that air
is withdrawn from the receiver, and that when it
descends the au: so withdrawn escapes i to the outer
atmosphere.
§ 1 16. To determine the density of the air in the
receiver after any number of strokes.
Let Fand v be the volumes of the receiver and
cylinder respectively. Let d be the original density
d by Google
The A ir-Pump, 1 43
of the air in r, when the piston is at b. Then Vd
equals the mass of air in r. When the piston rises to
A the mass of air in r occupies the space V-^-v ; and
if di be its diminished density we have
r^=(r+e/)//„or//, = y^ • ^,
since the mass remains the same.
When the piston descends to b a part of the air
escapes, and the mass of air in r = F//i. If now
d.2 represent the density of the air when its volume
increases to V + v, we have
Vd,={V+v) d,, or d, = {y^y d,
and similarly if d^ represent the density of the air m
R after three complete strokes —
Vd^ =
,^{V-^v) d,, or d, = {jr^' d,
and, consequently if d^ be the density of the air in
the receiver after n complete or double strokes
It will be seen that no amount of exhaustion can re-
duce the density of the air to zero. At each stroke
of the piston a fraction only of the air in the receiver
is removed ; and as the remaining air occupies the
whole volume of the receiver, a perfect vacuum can-
not possibly be obtained.
§ 117. Examples.— ( I.) If the volume of the receiver is
64 cubic inches and of the cylinder 8 cubic inches, what quan-
Digitized by VjOOQ IC
144 Pneumatics.
tity of air would be left in the receiver after two complete
strokes ?
At the first double stroke 8 cubic inches of air at the original
density would be removed and 56 would remain. Of these one-
eighth, or 7 cubic inches, would be removed at the second stroke,
and consequently 49 would be left.
(2.) After four complete strokes the density of the air in the
receiver is to its original density as 10,000 : 14,641 : compare the
volumes of the receiver and cylinder.
d 14,641 \K + z//
F+v V 14,641 II
§ 118. Difficulty of Workiiig.— It is to be ob-
served that the difficulty of working this kind of air-
pump increases with the number of strokes. For,
neglecting the frictional resistances^ the force required
to raise the piston depends on the difference of pres-
sure at ♦the upper and lower surfaces of the piston.
Now, this difference increases with the rarefaction of
the air in the receiver, and consequently the difficulty
of working the machine increases as the exhaustion
proceeds.
In lowering the piston the external pressure as-
sists the action, so long as the air in c is of less den-
sity than that outside ; but as soon as that point has
been reached an expenditure of force is necessary to
overcome the inside pressure and open the valve m.
Thus the difficulty of lowering the piston decreases
somewhat with the number of strokes.
§ 119. The Double-barrelled Air-Pump. — This
instrument, known as Hawksbee's air-pump, has two
Digitized by VjOOQ IC
The Double-barrelled A ir-pump. 145
Fig. (>t.
cylinders, in each of which is a piston worked by
a rack and pinion. To the
centre of the toothed-wheel
is fixed a handle, by moving
which to and fro the pistons
are made alternately to as-
cend and descend.
The chief advantage of
this instrument over the one
already described . is that a
volume of air equal to that of
the cylinder is removed at each single stroke of the
piston, and that, consequently, the rate of exhausting
the receiver is doubled.
In the adjoining figure the piston a is ascending
and withdrawing air from the receiver, and the piston
B is descending and discharging into the outer atmo-
sphere the air previously withdrawn from the receiver.
The position of the valves should be carefully noted.
Another advantage of this machine is that it is
easier to work. The difficulty of raising one piston
is compensated by the assistance which the pressure
of the atmosphere affords in forcing down the other
piston, and thus the difficulty of working the machine
does not increase, as in the case of the single-barrelled
air-pump, with the number of strokes. In pressing
down the piston no force is required till the air beneath
the piston has been compressed to the density of the
air outside ; and this occurs nearer and nearer to the
bottom of the cylinder as the degree of exhaustion
increases. Consequently, the instrument is worked
somewhat more easily as exhaustion proceeds.
Digitized by VjOOQ IC
146
Pnmmatics.
§ 120. Tate's Air-Pump. — This instrument combines
the advantage of double action with a single barrel.
The barrel is generally horizontal, and communicates
Fig. 67.
with the receiver by a vertical opening, o. The pis-
ton, c D, occupies a little less than half the length of
the barrel, and consists generally of two discs rigidly
connected together by the piston-rod which unites
both.
The principle of the action would be the same if
the piston were solid and of uniform area through-
out ; but the trouble <A working it would be greater.
The barrel is furnished with two valves, a and b, at
either end of it, opening outwards.
The action may be thus explained : —
Suppose the piston to be, first of all, in the posi-
tion shown in fig. i, all the air in firont of it having
been expelled through a by the driving of the pis-
ton home. If the piston be now pulled out, as in
Digitized by VjOOQ IC
Tate's Air-pump,
H7
fig. 2, the valve a will close, and the air in front of
the piston will pass through b, which will remain open.
When the piston reaches the farther end, b, as shown
in fig. 3, air will escape from the receiver through
o into the empty space left behind the piston. When
the piston is pushed inwards this air will be expelled
through A ; and on the piston reaching a, as shown in
fig. I, the empty space behind it will be again occupied
by the air firom the receiver. In this way, a certain
volume of air, equal to about half the contents of the
barrel, will be removed at each stroke of the piston ;
and, as the difference of pressure at the two ends of
the piston decreases with every stroke, the working
of the pump becomes easier as the exhaustion pro-
ceeds.
§ 121. Mercury Chmge. — The pressure of the air,
after any number of strokes, in the receiver of an air-
pump is indicated by a gauge, which is fig. 68.
generally attached to the connecting- =
pipe of the air-pump.
In its simplest form it consists of a
straight tube, open at both ends. The
upper end is connected with the receiver,
and the lower end dips into a cup of
mercury. As the air is removed firom
the receiver the pressure inside the tube
is less than that on the mercury in the
cup, and consequently the mercury rises ""
in the tube. This instrument enables us to watch the
process of exhaustion from the very first stroke. If
the barometer at the time marks 30 inches, and the
mercury has risen 4 inches in the tube, the pressure
L 2
Digitized by VjOOQ IC
14^
Pneumatics.
Fig. 69.
of the air in the receiver is 30— 4 = i6 in. The
necessary length of this form of gauge renders it
somewhat inconvenient.
The more commonly employed gauge consists of a
bent tube (fig. 69), having one end closed and the other
open. Each branch is about
ten inches in length, and the
closed end is filled with mer-
cury, the weight of which is
supported by the atmospheric
pressure. The instrument is
I gg II enclosed in a glass case, which
_!lL I I communicates with the re-
ceiver of the air-pump. The
first few strokes do not produce
any change in the gauge, but as
soon as the tension of the au: in the receiver is less
than the pressure due to the column of mercury in the
closed end of the tube the mercury begins to fall, and
the difference of level of the mercury in the two ends
measures the pressure of the air in the receiver. The
open end is SQmetimes bent again upon itself, as in
fig. 70, and screwed into the connecting pipe.
§ 122. Experiments. — ^The experiments that can
be performed with the air-pump are very numerous.
We have found it necessary to refer to some of them,
in order to prove that the air has weight (§ 90). The
following additional experiment should also be per-
formed : — Take a hollow cylindrical vessel and stretch
a bladder over one end, and then place it on the
plate of the air-pump, having carefully greased the
^dges of the glass vessel, so as to prevent the entrance
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Magdeburg Hemispheres.
149
of the air. As the air from ^ under the bladder is
gradually removed the bladder will be found to yield
under the pressure of the external air, and will at
length break.
By means of the Magdeburg hemispheres, in-
vented by Guericke, some idea may be formed of the
magnitude of the pressure which the atmosphere
exerts on a comparatively small surface. Their gene-
FlG. 71.
ral form is shown in fig. 71. The edges are greased
and pressed together, and the enclosed air is then
removed by the air-pump. In order to separate the
two hemispheres by pulling them asunder, a very
great effort must be made ; and this will be found to
be the case in whatever position the hemispheres may
be held, showing also that the atmospheric pressure
acts equally in all directions.
d by Google
150 Ptieiimatics,
§ 123. Sprengers Air-Pump. — By means of the
following contrivance a small receiver can be more
effectually exhausted than by any of the pumps
already described.
It consists of a tube, f b, open at both ends, and
fitted to a funnel a, by a piece of indiarubber. The
funnel contains mercury, the flow of
Fig. 72- which through the tube can be regu-
^Sil lated by tightening the indiarubber
▼^ connection. The tube is consider-
ably longer than a barometer tube,
and has a spout in its side attached
to the receiver, e, which is to be ex-
hausted. The lower end of the tube,
B, dips into a vessel of mercury. As
soon as the mercury begins to flow ex-
haustion commences. The first drops
of mercury that run out close the
A lower end of the tube and prevent
B M. the air from entering. As each drop
fiy^^^ of mercury passes the neck, c, the
jr|WS|^. air in the receiver, e, expands and
^''^^^^^^ occupies the space cq. In this
way the tube becomes filled with cylinders of air and
mercury, which gradually escape firom the spout into
the vessel of mercury. The mercury is poured again
into the funnel, a, and the process is repeated till the
tube is occupied by a continuous column of mercury,
the height of which is equal to that of the mercury
barometer. The exhaustion is now complete, and
the receiver, e, corresponds to the Torricellian vacuum
gf the Qrdinary barometer.
Digitized by VjOOQ IC
Sprengers A ir-pump, 151
§ 134* ^^ Condensing S]rrmge.-— This consists of
a barrel, a b, into which an air-tight piston fits, having
in it a valve that opens downwards. At the bottom
of the barrel is a valve that likewise opens
downwards and communicates with a re- ^^°* ^3-
ceiver, to which the syringe is tightly
screwed. When the piston is moved
down, the valve b is closed and the valve
A opened by the increased pressure. As
the piston returns, the pressure of the air
in the receiver closes the valve a, and
thus prevents the air from re-entering the
barrel It is evident that the same
quantity of air will enter the receiver at
each stroke of the piston, if the machine works per-
fectly and the piston is moved through the whole
length of the barrel.
Exercises. XII.
1. If the volume of the receiver of an air-pump be eight
times that of the barrel, compare the density of the air after the
third stroke with its original density.
2. If the receiver of an air-pump holds 90 grains of air at
the ordinary pressure, and if the barrel can hold 10 grains, what
will be the weight of the air in the receiver after four complete
strokes ?
3. If one-third of the contents of the receiver is removed at
each complete stroke, and the barometer stand at 75 cm., find
the height to which the mercury will rise in the simple barometer
gauge after three strokes.
4. The mercury rises in the barometer gauge through 2^ in.
in two complete strokes : compare the size of the barrel with
that of the receiver {h = 30).
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152 Pneumatics,
5. The brianches of a siphon barometer gauge are each
16 cm. long, and the closed branch is filled with mercury,
which also occupies I cm . of the open branch : compare the den-
sity of the air in the receiver with its original density, when
the mercury begins to fall in the siphon-gauge (^ = 75 cm.).
6. A receiver attached to an air-pump has the volume of
100 cubic inches, while the cylinder has the volume of 10 cubic
inches. What proportion of the original air will be left in the
receiver after the completion of the fourth double stroke?
7. The capacity of the barrel of a condensing air-pump is
10 cubic inches, and of a copper receiver 100 cubic inches. By
how much will the pressure of the air in the receiver be in-
creased after 20 strokes of the piston ?
8. When the height of the barometer is 75 cm.^ the air in
the receiver of an air-pump is exhausted until the mercury in
the barometer-gauge (fig. 68) attached to it rises from o to
36 cm. By how much has the tension of the enclosed air been
reduced ?
XIX. Pumps for Liquids,
% 125. Common or Suction Pump.— This is an
instrument for drawing water from a well or subter-
ranean reservoir.
It consists of a cylinder fitted with a piston and
valve, and connected by a second valve with a pipe
which communicates with the reservoir. The cylin-
der M N is called the pump-barrel ; the tube v e the
suction-tube. The mode of action is as follows : —
Suppose the piston at first to be at n, and the
suction-tube filled with air at the atmospheric pres-
sure. If the piston be raised the air in n d will ex-
pand, open the valve v, and follow the piston. At the
same time, since the pressure on the surface of the water
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The Cofftmon Pump.
153
Fig. 74.
Within the tube is diminished, ovAng to the expansion
of the contained air, the external pressure at d will
cause the water to rise in e to such a
height that the elasticity of the air below
A B, together with the weight of the column
of water above D in the suction-tube^
equals the ainiospheric pressure without.
As the piston descends the air below
it is compressed, and escapes after a
time through the valve f, the valve v
being closed by the increased pressure
of the air above it. Thus the water
remains at the same level in the suction-
tube whilst the piston is descending.
When the piston is again raised the
pressure is removed from above v, and
the air underneath it at once opens the
valve and occupies the space beneath
A B, the water rising in the suction-tube as before.
This action continues till the water has risen to n,
when the raising of the piston causes the water to
enter the barrel, provided the height d n is less than
Ji^y the height of the water barometer. As the piston
continues to rise the water will follow it so long as
the height of the piston above the level of the water
outside is less than H,
As the piston descends, the pressure of the water
beneath it opens the valve f, and the piston passes
through the water. When the piston again ascends
the water is discharged at the spout, and the barrel is
refilled through the suction-tube.
The water having once entered the barrel, the
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154 Pnmmatics.
contents of the barrel are discharged at each upward
stroke of the piston \ but, in order that a volume of
water equal to that of the barrel may be discharged
at each upward stroke, it is necessary that the atmos-
pheric pressure should first raise the water to the
spout ; — i,e, the height of the spout above the level of
the water outside must be less than H,
It is e\4dent that if the piston in descending does
not reach v, so as ultimately to exclude all air from
underneath it, the water may never be able to enter
the pump-barrel, even if n d is less than H,
It is not essential to the working of the pump
that the tube should be straight ; nor does it matter
at what horizontal distance from the pump-barrel the
suction-tube enters the reservoir.
§ 126. Force required to raise the piston-rod.
First Suppose the water has not yet entered the
barrel. In this case the force necessary to raise the
piston-rod is equal to the difference of the pressure
of the air on the top and bottom of the piston a b
(fig. 74). Let E represent the pressure of the air be-
low A B measured in water, and let ZTbe the height
of the water barometer outside. Let z equal the
height of the column of water in the tube.
Then H^=^ E •\- z\ and the force required to mise
the piston is
A B X {H— -^) = A B v. z
= the weight of a column of water that has a b for
base, and the distance of surface-level of water in the
pump above the level outside for height.
Secondly, Suppose the barrel abeady fiill of water.
The force required is, as before, the difference of pres-
sure on the two sides of the piston- r- i
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The Common Pump. 155
Let jp = c A (fig. 75) be tlie height of the water
above the piston at any instant ; then the pressure on
the upper surface of the piston is p^^
(Zr4-^)xAB
and the pressure on the lower surface is
(ZT— <s;)x A B
where z is, as before, the height of the water
below the piston above the level of the
reservoir. Hence, the difference, or force
required is
AB X {x->rz\
or the weight of the column of water having
A B for base and the difference of the level of the water
in the pump and in the reservoir for height. This,
therefore, is the measure of the tension of the piston-
rod in all cases.
When the pump is in full action, discharging at
each stroke a volume of water equal to that of the
barrel, the tension of the piston-rod is constant
§ 127. Ezamples. (i.) The length of the suction-tube of a
common pump is 12 feet, and the piston when at its lowest point
is 2 feet from the fixed valve ; if at the first stroke the water
rises 1 1 feet in the tube, find the extreme length of the stroke,
supposing the water barometer to stand at 33 feet, and the area
of the barrel to be three times that of the tube.
If jf be the length of the stroke, and a the area of the tube,
the volume originally occupied by the air is 12a + 2 x 3^ = i8<j ;
the volume occupied by the air after the first stroke is
3a (Jf + 2) + «=(3*+7) a.
,-. by Boyle's law, -il- «33zJ[i=?.
3^+7 33 3
or% Jf=6|feet.
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1 56 Pneumatics,
(2.) The suction-tube of a common'pump is 12 feet long, and
the piston, starting from the fixed valve, is raised at the first
stroke through 3 feet. If the area of the barrel is four times that
of the tube, find the height to which the water will be raised in
the tube {H = 34 feet).
Let X be the required height, a the area of suction-tube.
The volume originally occupied by the air was 12a, After the
first stroke the air occupies (12— jr) a + 3 x 4/7 = (24— jr) a ; and
the pressure is reduced from 34 to 34— ^ feet.
Hence, by Boyle's law.
12
. 5£_f , or jc « 8*2 feet nearly
M-x 34
§ 128. The Liftmg Pomp. — This is a modification
of the common pump, in which the water discharged
from the pump-barrel, which is closed at the top, en-
ters a pipe furnished with a valve and communicating
with the spout.
The action is the same as before ; but the water,
instead of flowing away through the spout, is lifted
into the pipe and prevented from returning by a valve.
In this way water can be stored up at any elevation,
and can afterwards be made to flow through the spout
as required.
§ 129. The Forcing Pump.— In this pump the
piston has no valve. The action is the same as in
Fig. 76. the common pump until the water enters
the barrel. Then, as the piston is raised
the water occupies the space beneath it,
and passing through the valve c, rises in
the tube c d to the same level as in the
barrel.
As the piston descends the valve b is
closed, and all the water contained in the
barrel {^forced up the pipe and prevented
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^ The Forcing Pump. 1 17
from returning by the valve C. In this way water
can be forced up to any height consistent with the
strength of the material, or can be made to nse m a
jet from the upper end of the pipe. No flow, how-
ever, occurs during the ascent of the piston.
§ 130. Forcing Pump, with Air-vessel.— To ob-
tain a continuous stream of water from the top of the
pipe c D (fig. 76) the water must be
first admitted into a strong vessel
containing air. The action being
the same as before, the piston in
descending forces the water into the
vessel E F (fig. 77), and compresses
the air in the upper part of it. As
more water is forced into the vessel
the air is further compressed, and
the water rises in the tube e d and flows out from d.
When the piston is drawn up the flow from d would
cease but for the fact that the air in e f, freed from its
former pressure, now expands and forces the water
up the tube, thus causing an unbroken flow from d.
The elasticity of the air in e f will decrease with
the escape of water from d, and consequently the
pressure of the water in the pipe must never be greater
than the excess of the pressure of the air in e f over
the ordinary atmospheric pressure. If the height of
the pipe is inconsiderable, the continuity of flow can
be easily preserved during the ascent of the piston.
§ 131. The Fire Engine. — This is a double forcing
pump connected with an air-chamber. The constancy
of flow is obtained not only by the air-vessel, but also
by the alternate action of the two pumps. The pistons
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158
Pneumatics.
are worked by a lever, so that one ascends while the
other descends. Every time the pistons momentarily
Fig. 78
Stop, the elasticity of the air in the air-chamber
maintains the flow.
§ 132. Brainah*8 Press. — This is a practical appli-
cation of Pascal's principle of the equal transmissi-
bility of fluid pressure, and consists of an apparatus
very similar to that explained in § 27, with the substi-
tution of a forcing-pump for the weighted piston.
In fig. 79, A is a platform which supports the
substance to be pressed against the strong framework
B. c and D are two solid cylinders which serve as
pistons, and e and f are two hollow cylinders con-
nected by a pipe furnished with a valve v.
The vessel f communicates with a reservoir of
water by the pipe h, so that d f h constitutes an ordi-
nary forcing pump. The piston d is attached to a
lever, k l m, and the power is applied at m.
When the instrument is in action the vessels e and
f are filled with water, and as the piston or plunger d
descends, it closes the valve z/ and forces the water
into the vessel e, and raises the cylinder c, with its
attached platform.
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BramaKs Press.
159
Fig. 79.
r
^
4- ^
1
\ c
^kjee!
IP
1
•IE «
If Q be the pressure produced by the plunger d in
its descent, this pressure is distributed equally over the
surface of the cylinder c ; and if A be the area of c,
A
and a that of d, then the pressure exerted at c is — C.
d
If we suppose P to be the force that must be applied
at M, to produce the pressure Q at l, then it follows
from the principle of the lever, that
Q\ P \\ MK : LK
or, G =
M K
iTk '
If, therefore, W represent the force with which the
cylinder, whose area is A^ is pressed upwards, we have
W^- . Q = ^ . ^. P; and if ^ and ^ are the
a a L K
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i6o Pneufnaiics,
radii of the two cylinders respectively, and M K = /
and L K = ^, we have
F f^ ' c
This instrument is very extensively employed
where the application of a considerable pressure
through a small space is required. The very great
pressure to which the water is subjected necessitates
great care in the construction of the collars through
which the cylinders work, so as to render them per-
fectly water-tight.
Exercises. XIIL
•
1. The spout of a common pump is i6 feet from the surface
of the water in the reservoir. The area of the pump-barrel is
72 square inches : find the tension of the piston-rod when the
pump is full of water.
2. The specific gravity of mercury is 13*6, and the height
of the mercurial barometer is 30 inches. What is the greatest
height to which water can be raised by means of the common
pump?
3. Find the pressure that can be produced by a Bramah's
press if the areas of the pistons are 8 : i, and if a force of
10 lbs. is applied at the end of a lever 2 feet long, and at a
distance of 20 ins. from the point where the piston-rod is at-
tached to it.
4. If in the lifting pump the area of the barrel is four times
as great as that of the pipe, compare the pressures on the two
valves at the commencement of the third stroke after the water
has entered the pipe.
5. If the area of the barrel of the forcing pump be 10 square
inches, and of the pipe into which the water is forced 2 Square
inches, find the height to which water can be raised in three com-
plete strokes, supposing the piston-range to be I foot
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Pneumatic Instruments. 1 6 1
6. Find the force that must be applied to the piston-rod, in
the preceding question, at the beginning of the third downward
stroke.
7. If the diameter of the piston of a common pum{> be
4 inches, and the height of the head of the water in the pump
18 feet above the well, find what pressure the piston bears,
taking the weight of i cubic foot of water equal to 62*3 lbs.
8. The length of the suction-tube is 20 feet, and the entire
stroke of the piston is 6 feet The piston starts from the bottom
of the barrel, and at the end of the first stroke the water has
risen 12 feet in the tube. Compare the area of the barrel with
that of the tube {H^ 34 feet).
9. A small strong pump is employed for raising mercury from
a vessel. The height of the fixed valve is 2 feet above the level
of the mercury in the vessel, and the spout is 8 in. above the
valve. When the pump is in full action what part of the con-
tents of the barrel can be ejected at each stroke of the piston
(^= 30 in. ) ?
10. A suction-tube is used for drawing up mercury from 3
vessel containing it, and the piston is raised 10 inches from its
lowest point, which is 2 inches from the valve at the bottom ol
the tube. To what height will the mercury rise in the tube
{iy=3oin.)?
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MISCELLANEOUS PROBLEMS.
1. A CUBE of brass, whose edge is 2 inches and specific
gravity 8, is completely imbedded in a cube of wood, whose
edge is 3 inches and specific gravity 5. Find the mean specific
gravity of the whole cube.
2. When equal volumes of alcohol (specific gravity «= o*8)
and distilled water are mixed together, the volume of the mix-
ture (after it has returned to its original temperature) is found
to fall short of the sum of the volumes of its coastituents by
4 per cent. Find the specific gravity of the mixture. (Univ.
Lond. Matric.)
3. The specific gravity of cast copper is 8*88, and that of
copper wire is 8*79. "What change of volume does a kilo-
gramme of cast copper undergo in being drawn out into wire?
(Univ. Lond., istB. Sc.)
4. A mixture is made of 7 cubic centimetres of sulphuric
acid (specific gravity = I '843^ and 3 cubic centimetres of dis-
tilled water, and its specific gravity when cold is found to be
1*615. Determine the contraction which has taken place.
(1st B. Sc. 1874.)
5. Two liquids are mixed (i) by volume in the proportion
of I : 4, arid (2) by weight in the proportion of 4 : I. The re-
sulting specific gravities are 2 and 3 respectively. Find the
specific gravities of the liquids.
Digitized by VjOOQ IC
Miscellaneous Problems. 163
6. The specific gravity of a mixture of two different liquids
being supposed to be an arithmetic mean between those of the
component liquids ; required the ratio of the volumes of the
latter contained in the mixture. (Univ. Lond., B.A. 1877.)
7. Find the pressure on a vertical rectangle, 10 inches long
and 6 inches broad, immersed in water with its longer sides
horizontal and with the upper one 2 inches below the surface.
(One cubic foot of water weighs i,ooo ounces.) (Matric. 1877.)
8. A vessel in the shape of a pyramid, 5 feet high, and with
a base 4 feet square, is filled with water. Find the pressure
upon the base, and account for its being greater than the total
weight in the vessel.
9. Find the whole pressure on the lower half of the curved
surface of a vertical cylinder filled with water, the area of the
base being I2| square cm. and the height 1*4 decim.
10. A piston, 6 square inches in area, is inserted into one
side of a closed cubical vessel, measuring 10 feet each way,
filled with water : the piston is pressed inwards with a force of
12 lbs. Find the increase of pressure produced on the entire
surface of the vessel. (B. A. 1872.)
11. The pressure at the bottom of a well is four times that
at the depth of 2 feet ; what is the depth of the well if the
pressure of the atmosphere is equivalent to 30 feet of water ?
(Camb. Gen. Exam. 1877.)
12. A and B are vessels fiiU of water, with circular and
horizontal bases, 12 inches and 8 inches in diameter respec-
tively. A is 8 inches, and B is 9 inches high. Compare the
pressure on the bases. (Cam. Gen. Exam.)
13. If a piece of wood weighing 120 lbs. floats in water with
four-fifths of its volume immersed, show what is its whole
volume (I cubic foot of water weighs 625 lbs.). (Matric. 1872.)
14. A wine-bottle, which below the neck is perfectly
M 2
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164 Hydrostatics and Pneumatics,
cylindrical and has a flat bottom, Is placed in pure water. It is
found to float upright, with 4J inches immersed. The bottle is
now removed from the wj^ter and put into oil, the specific
gravity of which is 0*915. How much of it will be immersed
in the latter fluid? (Matric. 1874.)
15. Two pieces of iron (specific gravity 77) suspended from
the two scale-pans of a balance, the one in water and the other
in alcohol of the specific gravity 0*85, are found to weigh
exactly alike. Find the proportion between their true weights.
(B. Sc. 1875.)
16. An inch cube of a substance of specific gravity i '2 is
immersed in a vessel containing two fluids which do not mix
The specific gravities of these fluids are i*o and i -5. Find what
will be the point at which the solid will rest. (Matric. 1876.)
17. A substance which weighs 14 lbs. in air and 12 lbs in
water, floats in mercury whose density is 13*6. What propor-
tion of its volume will be immersed ?
18. A solid, of which the volume is 1*6 cubic centimetres,
weighs 3*4 gprams in a fluid of specific gravity 0*85. Find the
specificgravity and weight of the substance. (B. Sc 1876).
19. An accurate balance is totally immersed in a vessel of
water. In one scale-pan some glass (specific gravity 2*5) is
being weighed, and exactly balances a one-pound weight
(specific gravity 8*o), which is placed in the other scale-pan.
Find the real weight of the glass. (Matric. 1875. )
20. A right cone, whose weight is W, floats in a liquid,
vortex downwards, with \ of its axis immersed ; what addi-
tional weight must be placed on the base of the cone so as just
to sink it entirely in the liquid. (Woolwich Exam.)
21. A cube floats in distilled water under the pressure of
the atmosphere, with four-fifthsof its volume immersed and with
two of its faces horizontal. When it is placed under a con-
denser where the pressure is that of ten atmospheres, find the
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Miscellaneous Problems. 165
alteration in the depth of immersion (the specific gravity of air
at the atmospheric pressure being; •cx)i3). (B. Sc.)
22. At the bottom of a mine a mercurial barometer stands at
77*4 centimetres ; what would be the height of an oil baro-
meter at the same place, the specific gravity of mercury being
13-596, and that of oil 09 ? (Matric. 1876.)
23. A certain quantity of air at atmospheric pressure has a
volume of 2 cubic feet, the temperature being 55° Fahr. What
does the volume of the air become when the pressure is increased
by one-twentieth, the temperature meanwhile remaining the
same? (Women*s Exam. 1876.)
24. A syphon barometer is so constructed that the long
closed tube has an internal sectional area equal to \ of an inch,
while the short open tube has an internal sectional area equal to
J an inch. Find what fall will take place in the long tube of
this barometer when the true pressure of the air falls one inch.
(B. Sc. 1875.)
25. The mercury in a barometer stands at £o inches ; the
section of the tube measures I square inch, and "the vacuum
above the mercury 6 cubic inches, as much air is passed up the
tube as depresses the mercury to 29 inches : what would be
the space occupied by the air under the atmospheric pressure ?
(B. Sc. 1872.)
26. In a tube of uniform bore a quantity of air is enclosed.
What will be the length of this column of air under a pressure
of three atmospheres, and what under a pressure of a third of an
atmosphere, its length under the pressure of a single atmosphere
being 12 inches? (B. A. 1876.)
27. A Marriotte's tube (fig. 51) has a uniform section of i
square inch, and is graduated in inches, 6 cubic inches are
enclosed in the shorter (closed) limb, when the mercury is at
the same level in both tubes. What volume of mercury must
be poured into the longer limb, in order to compress the air
Digitized by VjOOQ IC
1 66 Hydrostatics and Pneumatics.
into 2 inches ? The barometer stands at 30 inches. (B. Sc.
1874.)
28. Two cubic centimetres of air are measured off at atmo-
spheric pressure. When introduced into the vacuum of a baro-
meter they depress the mercury which previously stood at
76 cm., and occupy a volume of 15 cubic centimetres. By how
much has the mercurial column been depressed? (Matric. 1878.)
29. A tumbler full of air is placed mouth downwards under
water, at such a depth that the surface of the water inside it is
at a depth of 25^^ feet. Compare the weight of a cubic inch of
air in the tumbler with that of a cubic inch of air outside — the
barometer standing at 30 inches, and the specific gravity of
mercury being 13-6.
30. The contents of the receiver of an air-pump is six times
that of the barrel. Find the elastic force of the air in the re-
ceiver at the end of the eighth stroke of the piston, when the
atmospheric pressure is 15 lbs. to the square inch. (B. A.
1872.)
Richard Clay ^ Sons, Limited^ London &• Bungay,
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